Why does std::getline() skip input after a formatted extraction?
up vote
79
down vote
favorite
I have the following piece of code that prompts the user for their name and state:
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::cin >> name && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
}
What I find is that the name has been successfully extracted, but not the state. Here is the input and resulting output:
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in "
Why has the name of the state been omitted from the output? I've given the proper input, but the code somehow ignores it. Why does this happen?
c++ input iostream istream c++-faq
add a comment |
up vote
79
down vote
favorite
I have the following piece of code that prompts the user for their name and state:
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::cin >> name && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
}
What I find is that the name has been successfully extracted, but not the state. Here is the input and resulting output:
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in "
Why has the name of the state been omitted from the output? I've given the proper input, but the code somehow ignores it. Why does this happen?
c++ input iostream istream c++-faq
I believestd::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).
– jww
Nov 11 at 17:30
add a comment |
up vote
79
down vote
favorite
up vote
79
down vote
favorite
I have the following piece of code that prompts the user for their name and state:
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::cin >> name && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
}
What I find is that the name has been successfully extracted, but not the state. Here is the input and resulting output:
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in "
Why has the name of the state been omitted from the output? I've given the proper input, but the code somehow ignores it. Why does this happen?
c++ input iostream istream c++-faq
I have the following piece of code that prompts the user for their name and state:
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::cin >> name && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
}
What I find is that the name has been successfully extracted, but not the state. Here is the input and resulting output:
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in "
Why has the name of the state been omitted from the output? I've given the proper input, but the code somehow ignores it. Why does this happen?
c++ input iostream istream c++-faq
c++ input iostream istream c++-faq
edited Dec 23 '16 at 16:02
πάντα ῥεῖ
71.4k972134
71.4k972134
asked Feb 5 '14 at 2:01
0x499602D2
66.9k26116201
66.9k26116201
I believestd::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).
– jww
Nov 11 at 17:30
add a comment |
I believestd::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).
– jww
Nov 11 at 17:30
I believe
std::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).– jww
Nov 11 at 17:30
I believe
std::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).– jww
Nov 11 at 17:30
add a comment |
3 Answers
3
active
oldest
votes
up vote
92
down vote
Why does this happen?
This has little to do with the input you provided yourself but rather with the default behavior std::getline()
exhibits. When you provided your input for the name (std::cin >> name
), you not only submitted the following characters, but also an implicit newline was appended to the stream:
"Johnn"
A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into name
until the next I/O operation where it is either discarded or consumed. When the flow of control reaches std::getline()
, the newline will be discarded, but the input will cease immediately. The reason this happens is because the default functionality of this function dictates that it should (it attempts to read a line and stops when it finds a newline).
Because this leading newline inhibits the expected functionality of your program, it follows that it must be skipped our ignored somehow. One option is to call std::cin.ignore()
after the the first extraction. It will discard the next available character so that the newline is no longer intrusive.
In-Depth Explanation:
This is the overload of std::getline()
that you called:
template<class charT>
std::basic_istream<charT>& getline( std::basic_istream<charT>& input,
std::basic_string<charT>& str )
Another overload of this function takes a delimiter of type charT
. A delimiter character is a character that represents the boundary between sequences of input. This particular overload sets the delimiter to the newline character input.widen('n')
by default since one was not supplied.
Now, these are a few of the conditions whereby std::getline()
terminates input:
- If the stream has extracted the maximum amount of characters a
std::basic_string<charT>
can hold - If the end-of-file (EOF) character has been found
- If the delimiter has been found
The third condition is the one we're dealing with. Your input into state
is represented thusly:
"JohnnNew Hampshire"
^
|
next_pointer
where next_pointer
is the next character to be parsed. Since the character stored at the next position in the input sequence is the delimiter, std::getline()
will quietly discard that character, increment next_pointer
to the next available character, and stop input. This means that the rest of the characters that you have provided still remain in the buffer for the next I/O operation. You'll notice that if you perform another read from the line into state
, your extraction will yield the correct result as the last call to std::getline()
discarded the delimiter.
You may have noticed that you don't typically run into this problem when extracting with the formatted input operator (operator>>()
). This is because input streams use whitespace as delimiters for input and have the std::skipws
1 manipulator set on by default. Streams will discard the leading whitespace from the stream when beginning to perform formatted input.2
Unlike the formatted input operators, std::getline()
is an unformatted input function. And all unformatted input functions have the following code somewhat in common:
typename std::basic_istream<charT>::sentry ok(istream_object, true);
The above is a sentry object which is instantiated in all formatted/unformatted I/O functions in a standard C++ implementation. Sentry objects are used for preparing the stream for I/O and determining whether or not it is in a fail state. You'll only find that in the unformatted input functions, the second argument to the sentry constructor is true
. That argument means that leading whitespace will not be discarded from the beginning of the input sequence. Here is the relevant quote from the Standard [§27.7.2.1.3/2]:
explicit sentry(basic_istream<charT, traits>& is, bool noskipws = false);
[...] If
noskipws
is zero andis.flags() & ios_base::skipws
is nonzero, the function extracts and discards each character as long as the next available input characterc
is a whitespace character. [...]
Since the above condition is false, the sentry object will not discard the whitespace. The reason noskipws
is set to true
by this function is because the point of std::getline()
is to read raw, unformatted characters into a std::basic_string<charT>
object.
The Solution:
There's no way to stop this behavior of std::getline()
. What you'll have to do is discard the new line yourself before std::getline()
runs (but do it after the formatted extraction). This can be done by using ignore()
to discard the rest of the input until we reach a fresh new line:
if (std::cin >> name &&
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), 'n') &&
std::getline(std::cin, state))
{ ... }
You'll need to include <limits>
to use std::numeric_limits
. std::basic_istream<...>::ignore()
is a function that discards a specified amount of characters until it either finds a delimiter or reaches the end of the stream (ignore()
also discards the delimiter if it finds it). The max()
function returns the largest amount of characters that a stream can accept.
Another way to discard the whitespace is to use the std::ws
function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:
if (std::cin >> name && std::getline(std::cin >> std::ws, state))
{ ... }
What's the difference?
The difference is that ignore(std::streamsize count = 1, int_type delim = Traits::eof())
3 indiscriminately discards characters until it either discards count
characters, finds the delimiter (specified by the second argument delim
) or hits the end of the stream. std::ws
is only used for discarding whitespace characters from the beginning of the stream.
If you are mixing formatted input with unformatted input and you need to discard residual whitespace, use std::ws
. Otherwise, if you need to clear out invalid input regardless of what it is, use ignore()
. In our example, we only need to clear whitespace since the stream consumed your input of "John"
for the name
variable. All that was left was the newline character.
1: std::skipws
is manipulator that tells the input stream to discard leading whitespace when performing formatted input. This can be turned off with the std::noskipws
manipulator.
2: Input streams deem certain characters as whitespace by default, such the space character, newline character, form feed, carriage return, etc.
3: This is the signature of std::basic_istream<...>::ignore()
. You can call it with zero arguments to discard a single character from the stream, one argument to discard a certain amount of characters, or two arguments to discard count
characters or until it reaches delim
, whichever one comes first. You normally use std::numeric_limits<std::streamsize>::max()
as the value of count
if you don't know how many characters there are before the delimiter, but you want to discard them anyway.
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then usestd::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just usestd::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.
– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
add a comment |
up vote
10
down vote
Everything will be OK if you change your initial code in the following way:
if ((cin >> name).get() && std::getline(cin, state))
2
Thank you. This will also work becauseget()
consumes the next character. There's also(std::cin >> name).ignore()
which I suggested earlier in my answer.
– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
add a comment |
up vote
0
down vote
This happens because an implicit line feed also known as newline character n
is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline
when checking for multiple lines of user input. The default behavior of std::getline
will read everything up to and including the newline character n
from the input stream object which is std::cin
in this case.
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::getline(std::cin, name) && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
return 0;
}
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in New Hampshire"
add a comment |
protected by StoryTeller Dec 6 '17 at 8:53
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
92
down vote
Why does this happen?
This has little to do with the input you provided yourself but rather with the default behavior std::getline()
exhibits. When you provided your input for the name (std::cin >> name
), you not only submitted the following characters, but also an implicit newline was appended to the stream:
"Johnn"
A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into name
until the next I/O operation where it is either discarded or consumed. When the flow of control reaches std::getline()
, the newline will be discarded, but the input will cease immediately. The reason this happens is because the default functionality of this function dictates that it should (it attempts to read a line and stops when it finds a newline).
Because this leading newline inhibits the expected functionality of your program, it follows that it must be skipped our ignored somehow. One option is to call std::cin.ignore()
after the the first extraction. It will discard the next available character so that the newline is no longer intrusive.
In-Depth Explanation:
This is the overload of std::getline()
that you called:
template<class charT>
std::basic_istream<charT>& getline( std::basic_istream<charT>& input,
std::basic_string<charT>& str )
Another overload of this function takes a delimiter of type charT
. A delimiter character is a character that represents the boundary between sequences of input. This particular overload sets the delimiter to the newline character input.widen('n')
by default since one was not supplied.
Now, these are a few of the conditions whereby std::getline()
terminates input:
- If the stream has extracted the maximum amount of characters a
std::basic_string<charT>
can hold - If the end-of-file (EOF) character has been found
- If the delimiter has been found
The third condition is the one we're dealing with. Your input into state
is represented thusly:
"JohnnNew Hampshire"
^
|
next_pointer
where next_pointer
is the next character to be parsed. Since the character stored at the next position in the input sequence is the delimiter, std::getline()
will quietly discard that character, increment next_pointer
to the next available character, and stop input. This means that the rest of the characters that you have provided still remain in the buffer for the next I/O operation. You'll notice that if you perform another read from the line into state
, your extraction will yield the correct result as the last call to std::getline()
discarded the delimiter.
You may have noticed that you don't typically run into this problem when extracting with the formatted input operator (operator>>()
). This is because input streams use whitespace as delimiters for input and have the std::skipws
1 manipulator set on by default. Streams will discard the leading whitespace from the stream when beginning to perform formatted input.2
Unlike the formatted input operators, std::getline()
is an unformatted input function. And all unformatted input functions have the following code somewhat in common:
typename std::basic_istream<charT>::sentry ok(istream_object, true);
The above is a sentry object which is instantiated in all formatted/unformatted I/O functions in a standard C++ implementation. Sentry objects are used for preparing the stream for I/O and determining whether or not it is in a fail state. You'll only find that in the unformatted input functions, the second argument to the sentry constructor is true
. That argument means that leading whitespace will not be discarded from the beginning of the input sequence. Here is the relevant quote from the Standard [§27.7.2.1.3/2]:
explicit sentry(basic_istream<charT, traits>& is, bool noskipws = false);
[...] If
noskipws
is zero andis.flags() & ios_base::skipws
is nonzero, the function extracts and discards each character as long as the next available input characterc
is a whitespace character. [...]
Since the above condition is false, the sentry object will not discard the whitespace. The reason noskipws
is set to true
by this function is because the point of std::getline()
is to read raw, unformatted characters into a std::basic_string<charT>
object.
The Solution:
There's no way to stop this behavior of std::getline()
. What you'll have to do is discard the new line yourself before std::getline()
runs (but do it after the formatted extraction). This can be done by using ignore()
to discard the rest of the input until we reach a fresh new line:
if (std::cin >> name &&
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), 'n') &&
std::getline(std::cin, state))
{ ... }
You'll need to include <limits>
to use std::numeric_limits
. std::basic_istream<...>::ignore()
is a function that discards a specified amount of characters until it either finds a delimiter or reaches the end of the stream (ignore()
also discards the delimiter if it finds it). The max()
function returns the largest amount of characters that a stream can accept.
Another way to discard the whitespace is to use the std::ws
function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:
if (std::cin >> name && std::getline(std::cin >> std::ws, state))
{ ... }
What's the difference?
The difference is that ignore(std::streamsize count = 1, int_type delim = Traits::eof())
3 indiscriminately discards characters until it either discards count
characters, finds the delimiter (specified by the second argument delim
) or hits the end of the stream. std::ws
is only used for discarding whitespace characters from the beginning of the stream.
If you are mixing formatted input with unformatted input and you need to discard residual whitespace, use std::ws
. Otherwise, if you need to clear out invalid input regardless of what it is, use ignore()
. In our example, we only need to clear whitespace since the stream consumed your input of "John"
for the name
variable. All that was left was the newline character.
1: std::skipws
is manipulator that tells the input stream to discard leading whitespace when performing formatted input. This can be turned off with the std::noskipws
manipulator.
2: Input streams deem certain characters as whitespace by default, such the space character, newline character, form feed, carriage return, etc.
3: This is the signature of std::basic_istream<...>::ignore()
. You can call it with zero arguments to discard a single character from the stream, one argument to discard a certain amount of characters, or two arguments to discard count
characters or until it reaches delim
, whichever one comes first. You normally use std::numeric_limits<std::streamsize>::max()
as the value of count
if you don't know how many characters there are before the delimiter, but you want to discard them anyway.
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then usestd::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just usestd::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.
– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
add a comment |
up vote
92
down vote
Why does this happen?
This has little to do with the input you provided yourself but rather with the default behavior std::getline()
exhibits. When you provided your input for the name (std::cin >> name
), you not only submitted the following characters, but also an implicit newline was appended to the stream:
"Johnn"
A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into name
until the next I/O operation where it is either discarded or consumed. When the flow of control reaches std::getline()
, the newline will be discarded, but the input will cease immediately. The reason this happens is because the default functionality of this function dictates that it should (it attempts to read a line and stops when it finds a newline).
Because this leading newline inhibits the expected functionality of your program, it follows that it must be skipped our ignored somehow. One option is to call std::cin.ignore()
after the the first extraction. It will discard the next available character so that the newline is no longer intrusive.
In-Depth Explanation:
This is the overload of std::getline()
that you called:
template<class charT>
std::basic_istream<charT>& getline( std::basic_istream<charT>& input,
std::basic_string<charT>& str )
Another overload of this function takes a delimiter of type charT
. A delimiter character is a character that represents the boundary between sequences of input. This particular overload sets the delimiter to the newline character input.widen('n')
by default since one was not supplied.
Now, these are a few of the conditions whereby std::getline()
terminates input:
- If the stream has extracted the maximum amount of characters a
std::basic_string<charT>
can hold - If the end-of-file (EOF) character has been found
- If the delimiter has been found
The third condition is the one we're dealing with. Your input into state
is represented thusly:
"JohnnNew Hampshire"
^
|
next_pointer
where next_pointer
is the next character to be parsed. Since the character stored at the next position in the input sequence is the delimiter, std::getline()
will quietly discard that character, increment next_pointer
to the next available character, and stop input. This means that the rest of the characters that you have provided still remain in the buffer for the next I/O operation. You'll notice that if you perform another read from the line into state
, your extraction will yield the correct result as the last call to std::getline()
discarded the delimiter.
You may have noticed that you don't typically run into this problem when extracting with the formatted input operator (operator>>()
). This is because input streams use whitespace as delimiters for input and have the std::skipws
1 manipulator set on by default. Streams will discard the leading whitespace from the stream when beginning to perform formatted input.2
Unlike the formatted input operators, std::getline()
is an unformatted input function. And all unformatted input functions have the following code somewhat in common:
typename std::basic_istream<charT>::sentry ok(istream_object, true);
The above is a sentry object which is instantiated in all formatted/unformatted I/O functions in a standard C++ implementation. Sentry objects are used for preparing the stream for I/O and determining whether or not it is in a fail state. You'll only find that in the unformatted input functions, the second argument to the sentry constructor is true
. That argument means that leading whitespace will not be discarded from the beginning of the input sequence. Here is the relevant quote from the Standard [§27.7.2.1.3/2]:
explicit sentry(basic_istream<charT, traits>& is, bool noskipws = false);
[...] If
noskipws
is zero andis.flags() & ios_base::skipws
is nonzero, the function extracts and discards each character as long as the next available input characterc
is a whitespace character. [...]
Since the above condition is false, the sentry object will not discard the whitespace. The reason noskipws
is set to true
by this function is because the point of std::getline()
is to read raw, unformatted characters into a std::basic_string<charT>
object.
The Solution:
There's no way to stop this behavior of std::getline()
. What you'll have to do is discard the new line yourself before std::getline()
runs (but do it after the formatted extraction). This can be done by using ignore()
to discard the rest of the input until we reach a fresh new line:
if (std::cin >> name &&
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), 'n') &&
std::getline(std::cin, state))
{ ... }
You'll need to include <limits>
to use std::numeric_limits
. std::basic_istream<...>::ignore()
is a function that discards a specified amount of characters until it either finds a delimiter or reaches the end of the stream (ignore()
also discards the delimiter if it finds it). The max()
function returns the largest amount of characters that a stream can accept.
Another way to discard the whitespace is to use the std::ws
function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:
if (std::cin >> name && std::getline(std::cin >> std::ws, state))
{ ... }
What's the difference?
The difference is that ignore(std::streamsize count = 1, int_type delim = Traits::eof())
3 indiscriminately discards characters until it either discards count
characters, finds the delimiter (specified by the second argument delim
) or hits the end of the stream. std::ws
is only used for discarding whitespace characters from the beginning of the stream.
If you are mixing formatted input with unformatted input and you need to discard residual whitespace, use std::ws
. Otherwise, if you need to clear out invalid input regardless of what it is, use ignore()
. In our example, we only need to clear whitespace since the stream consumed your input of "John"
for the name
variable. All that was left was the newline character.
1: std::skipws
is manipulator that tells the input stream to discard leading whitespace when performing formatted input. This can be turned off with the std::noskipws
manipulator.
2: Input streams deem certain characters as whitespace by default, such the space character, newline character, form feed, carriage return, etc.
3: This is the signature of std::basic_istream<...>::ignore()
. You can call it with zero arguments to discard a single character from the stream, one argument to discard a certain amount of characters, or two arguments to discard count
characters or until it reaches delim
, whichever one comes first. You normally use std::numeric_limits<std::streamsize>::max()
as the value of count
if you don't know how many characters there are before the delimiter, but you want to discard them anyway.
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then usestd::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just usestd::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.
– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
add a comment |
up vote
92
down vote
up vote
92
down vote
Why does this happen?
This has little to do with the input you provided yourself but rather with the default behavior std::getline()
exhibits. When you provided your input for the name (std::cin >> name
), you not only submitted the following characters, but also an implicit newline was appended to the stream:
"Johnn"
A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into name
until the next I/O operation where it is either discarded or consumed. When the flow of control reaches std::getline()
, the newline will be discarded, but the input will cease immediately. The reason this happens is because the default functionality of this function dictates that it should (it attempts to read a line and stops when it finds a newline).
Because this leading newline inhibits the expected functionality of your program, it follows that it must be skipped our ignored somehow. One option is to call std::cin.ignore()
after the the first extraction. It will discard the next available character so that the newline is no longer intrusive.
In-Depth Explanation:
This is the overload of std::getline()
that you called:
template<class charT>
std::basic_istream<charT>& getline( std::basic_istream<charT>& input,
std::basic_string<charT>& str )
Another overload of this function takes a delimiter of type charT
. A delimiter character is a character that represents the boundary between sequences of input. This particular overload sets the delimiter to the newline character input.widen('n')
by default since one was not supplied.
Now, these are a few of the conditions whereby std::getline()
terminates input:
- If the stream has extracted the maximum amount of characters a
std::basic_string<charT>
can hold - If the end-of-file (EOF) character has been found
- If the delimiter has been found
The third condition is the one we're dealing with. Your input into state
is represented thusly:
"JohnnNew Hampshire"
^
|
next_pointer
where next_pointer
is the next character to be parsed. Since the character stored at the next position in the input sequence is the delimiter, std::getline()
will quietly discard that character, increment next_pointer
to the next available character, and stop input. This means that the rest of the characters that you have provided still remain in the buffer for the next I/O operation. You'll notice that if you perform another read from the line into state
, your extraction will yield the correct result as the last call to std::getline()
discarded the delimiter.
You may have noticed that you don't typically run into this problem when extracting with the formatted input operator (operator>>()
). This is because input streams use whitespace as delimiters for input and have the std::skipws
1 manipulator set on by default. Streams will discard the leading whitespace from the stream when beginning to perform formatted input.2
Unlike the formatted input operators, std::getline()
is an unformatted input function. And all unformatted input functions have the following code somewhat in common:
typename std::basic_istream<charT>::sentry ok(istream_object, true);
The above is a sentry object which is instantiated in all formatted/unformatted I/O functions in a standard C++ implementation. Sentry objects are used for preparing the stream for I/O and determining whether or not it is in a fail state. You'll only find that in the unformatted input functions, the second argument to the sentry constructor is true
. That argument means that leading whitespace will not be discarded from the beginning of the input sequence. Here is the relevant quote from the Standard [§27.7.2.1.3/2]:
explicit sentry(basic_istream<charT, traits>& is, bool noskipws = false);
[...] If
noskipws
is zero andis.flags() & ios_base::skipws
is nonzero, the function extracts and discards each character as long as the next available input characterc
is a whitespace character. [...]
Since the above condition is false, the sentry object will not discard the whitespace. The reason noskipws
is set to true
by this function is because the point of std::getline()
is to read raw, unformatted characters into a std::basic_string<charT>
object.
The Solution:
There's no way to stop this behavior of std::getline()
. What you'll have to do is discard the new line yourself before std::getline()
runs (but do it after the formatted extraction). This can be done by using ignore()
to discard the rest of the input until we reach a fresh new line:
if (std::cin >> name &&
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), 'n') &&
std::getline(std::cin, state))
{ ... }
You'll need to include <limits>
to use std::numeric_limits
. std::basic_istream<...>::ignore()
is a function that discards a specified amount of characters until it either finds a delimiter or reaches the end of the stream (ignore()
also discards the delimiter if it finds it). The max()
function returns the largest amount of characters that a stream can accept.
Another way to discard the whitespace is to use the std::ws
function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:
if (std::cin >> name && std::getline(std::cin >> std::ws, state))
{ ... }
What's the difference?
The difference is that ignore(std::streamsize count = 1, int_type delim = Traits::eof())
3 indiscriminately discards characters until it either discards count
characters, finds the delimiter (specified by the second argument delim
) or hits the end of the stream. std::ws
is only used for discarding whitespace characters from the beginning of the stream.
If you are mixing formatted input with unformatted input and you need to discard residual whitespace, use std::ws
. Otherwise, if you need to clear out invalid input regardless of what it is, use ignore()
. In our example, we only need to clear whitespace since the stream consumed your input of "John"
for the name
variable. All that was left was the newline character.
1: std::skipws
is manipulator that tells the input stream to discard leading whitespace when performing formatted input. This can be turned off with the std::noskipws
manipulator.
2: Input streams deem certain characters as whitespace by default, such the space character, newline character, form feed, carriage return, etc.
3: This is the signature of std::basic_istream<...>::ignore()
. You can call it with zero arguments to discard a single character from the stream, one argument to discard a certain amount of characters, or two arguments to discard count
characters or until it reaches delim
, whichever one comes first. You normally use std::numeric_limits<std::streamsize>::max()
as the value of count
if you don't know how many characters there are before the delimiter, but you want to discard them anyway.
Why does this happen?
This has little to do with the input you provided yourself but rather with the default behavior std::getline()
exhibits. When you provided your input for the name (std::cin >> name
), you not only submitted the following characters, but also an implicit newline was appended to the stream:
"Johnn"
A newline is always appended to your input when you select Enter or Return when submitting from a terminal. It is also used in files for moving toward the next line. The newline is left in the buffer after the extraction into name
until the next I/O operation where it is either discarded or consumed. When the flow of control reaches std::getline()
, the newline will be discarded, but the input will cease immediately. The reason this happens is because the default functionality of this function dictates that it should (it attempts to read a line and stops when it finds a newline).
Because this leading newline inhibits the expected functionality of your program, it follows that it must be skipped our ignored somehow. One option is to call std::cin.ignore()
after the the first extraction. It will discard the next available character so that the newline is no longer intrusive.
In-Depth Explanation:
This is the overload of std::getline()
that you called:
template<class charT>
std::basic_istream<charT>& getline( std::basic_istream<charT>& input,
std::basic_string<charT>& str )
Another overload of this function takes a delimiter of type charT
. A delimiter character is a character that represents the boundary between sequences of input. This particular overload sets the delimiter to the newline character input.widen('n')
by default since one was not supplied.
Now, these are a few of the conditions whereby std::getline()
terminates input:
- If the stream has extracted the maximum amount of characters a
std::basic_string<charT>
can hold - If the end-of-file (EOF) character has been found
- If the delimiter has been found
The third condition is the one we're dealing with. Your input into state
is represented thusly:
"JohnnNew Hampshire"
^
|
next_pointer
where next_pointer
is the next character to be parsed. Since the character stored at the next position in the input sequence is the delimiter, std::getline()
will quietly discard that character, increment next_pointer
to the next available character, and stop input. This means that the rest of the characters that you have provided still remain in the buffer for the next I/O operation. You'll notice that if you perform another read from the line into state
, your extraction will yield the correct result as the last call to std::getline()
discarded the delimiter.
You may have noticed that you don't typically run into this problem when extracting with the formatted input operator (operator>>()
). This is because input streams use whitespace as delimiters for input and have the std::skipws
1 manipulator set on by default. Streams will discard the leading whitespace from the stream when beginning to perform formatted input.2
Unlike the formatted input operators, std::getline()
is an unformatted input function. And all unformatted input functions have the following code somewhat in common:
typename std::basic_istream<charT>::sentry ok(istream_object, true);
The above is a sentry object which is instantiated in all formatted/unformatted I/O functions in a standard C++ implementation. Sentry objects are used for preparing the stream for I/O and determining whether or not it is in a fail state. You'll only find that in the unformatted input functions, the second argument to the sentry constructor is true
. That argument means that leading whitespace will not be discarded from the beginning of the input sequence. Here is the relevant quote from the Standard [§27.7.2.1.3/2]:
explicit sentry(basic_istream<charT, traits>& is, bool noskipws = false);
[...] If
noskipws
is zero andis.flags() & ios_base::skipws
is nonzero, the function extracts and discards each character as long as the next available input characterc
is a whitespace character. [...]
Since the above condition is false, the sentry object will not discard the whitespace. The reason noskipws
is set to true
by this function is because the point of std::getline()
is to read raw, unformatted characters into a std::basic_string<charT>
object.
The Solution:
There's no way to stop this behavior of std::getline()
. What you'll have to do is discard the new line yourself before std::getline()
runs (but do it after the formatted extraction). This can be done by using ignore()
to discard the rest of the input until we reach a fresh new line:
if (std::cin >> name &&
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), 'n') &&
std::getline(std::cin, state))
{ ... }
You'll need to include <limits>
to use std::numeric_limits
. std::basic_istream<...>::ignore()
is a function that discards a specified amount of characters until it either finds a delimiter or reaches the end of the stream (ignore()
also discards the delimiter if it finds it). The max()
function returns the largest amount of characters that a stream can accept.
Another way to discard the whitespace is to use the std::ws
function which is a manipulator designed to extract and discard leading whitespace from the beginning of an input stream:
if (std::cin >> name && std::getline(std::cin >> std::ws, state))
{ ... }
What's the difference?
The difference is that ignore(std::streamsize count = 1, int_type delim = Traits::eof())
3 indiscriminately discards characters until it either discards count
characters, finds the delimiter (specified by the second argument delim
) or hits the end of the stream. std::ws
is only used for discarding whitespace characters from the beginning of the stream.
If you are mixing formatted input with unformatted input and you need to discard residual whitespace, use std::ws
. Otherwise, if you need to clear out invalid input regardless of what it is, use ignore()
. In our example, we only need to clear whitespace since the stream consumed your input of "John"
for the name
variable. All that was left was the newline character.
1: std::skipws
is manipulator that tells the input stream to discard leading whitespace when performing formatted input. This can be turned off with the std::noskipws
manipulator.
2: Input streams deem certain characters as whitespace by default, such the space character, newline character, form feed, carriage return, etc.
3: This is the signature of std::basic_istream<...>::ignore()
. You can call it with zero arguments to discard a single character from the stream, one argument to discard a certain amount of characters, or two arguments to discard count
characters or until it reaches delim
, whichever one comes first. You normally use std::numeric_limits<std::streamsize>::max()
as the value of count
if you don't know how many characters there are before the delimiter, but you want to discard them anyway.
edited Aug 2 '16 at 3:56
user -1
456
456
answered Feb 5 '14 at 2:01
0x499602D2
66.9k26116201
66.9k26116201
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then usestd::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just usestd::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.
– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
add a comment |
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then usestd::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just usestd::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.
– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
Why not simply
if (getline(std::cin, name) && getline(std::cin, state))
?– Fred Larson
Aug 19 '16 at 19:41
Why not simply
if (getline(std::cin, name) && getline(std::cin, state))
?– Fred Larson
Aug 19 '16 at 19:41
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
@FredLarson Good point. Though it wouldn't work if the first extraction is of an integer or anything that isn't a string.
– 0x499602D2
Aug 19 '16 at 20:30
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then use
std::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just use std::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.– Fred Larson
Aug 19 '16 at 20:35
Of course, that isn't the case here and there's no point in doing the same thing two different ways. For an integer you could get the line into a string and then use
std::stoi()
, but then it's not so clear there's an advantage. But I tend to prefer to just use std::getline()
for line-oriented input and then deal with parsing the line in whatever way makes sense. I think it's less error prone.– Fred Larson
Aug 19 '16 at 20:35
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
@FredLarson Agreed. Maybe I'll add that in if I have the time.
– 0x499602D2
Aug 19 '16 at 20:39
add a comment |
up vote
10
down vote
Everything will be OK if you change your initial code in the following way:
if ((cin >> name).get() && std::getline(cin, state))
2
Thank you. This will also work becauseget()
consumes the next character. There's also(std::cin >> name).ignore()
which I suggested earlier in my answer.
– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
add a comment |
up vote
10
down vote
Everything will be OK if you change your initial code in the following way:
if ((cin >> name).get() && std::getline(cin, state))
2
Thank you. This will also work becauseget()
consumes the next character. There's also(std::cin >> name).ignore()
which I suggested earlier in my answer.
– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
add a comment |
up vote
10
down vote
up vote
10
down vote
Everything will be OK if you change your initial code in the following way:
if ((cin >> name).get() && std::getline(cin, state))
Everything will be OK if you change your initial code in the following way:
if ((cin >> name).get() && std::getline(cin, state))
edited Mar 26 '14 at 12:21
Qantas 94 Heavy
11.9k155370
11.9k155370
answered Mar 26 '14 at 12:01
Boris
1573
1573
2
Thank you. This will also work becauseget()
consumes the next character. There's also(std::cin >> name).ignore()
which I suggested earlier in my answer.
– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
add a comment |
2
Thank you. This will also work becauseget()
consumes the next character. There's also(std::cin >> name).ignore()
which I suggested earlier in my answer.
– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
Why not simplyif (getline(std::cin, name) && getline(std::cin, state))
?
– Fred Larson
Aug 19 '16 at 19:41
2
2
Thank you. This will also work because
get()
consumes the next character. There's also (std::cin >> name).ignore()
which I suggested earlier in my answer.– 0x499602D2
Mar 26 '14 at 12:14
Thank you. This will also work because
get()
consumes the next character. There's also (std::cin >> name).ignore()
which I suggested earlier in my answer.– 0x499602D2
Mar 26 '14 at 12:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
"..work because get()..." Yes, exactly. Sorry for giving the answer without details.
– Boris
Mar 26 '14 at 13:14
4
4
Why not simply
if (getline(std::cin, name) && getline(std::cin, state))
?– Fred Larson
Aug 19 '16 at 19:41
Why not simply
if (getline(std::cin, name) && getline(std::cin, state))
?– Fred Larson
Aug 19 '16 at 19:41
add a comment |
up vote
0
down vote
This happens because an implicit line feed also known as newline character n
is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline
when checking for multiple lines of user input. The default behavior of std::getline
will read everything up to and including the newline character n
from the input stream object which is std::cin
in this case.
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::getline(std::cin, name) && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
return 0;
}
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in New Hampshire"
add a comment |
up vote
0
down vote
This happens because an implicit line feed also known as newline character n
is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline
when checking for multiple lines of user input. The default behavior of std::getline
will read everything up to and including the newline character n
from the input stream object which is std::cin
in this case.
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::getline(std::cin, name) && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
return 0;
}
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in New Hampshire"
add a comment |
up vote
0
down vote
up vote
0
down vote
This happens because an implicit line feed also known as newline character n
is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline
when checking for multiple lines of user input. The default behavior of std::getline
will read everything up to and including the newline character n
from the input stream object which is std::cin
in this case.
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::getline(std::cin, name) && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
return 0;
}
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in New Hampshire"
This happens because an implicit line feed also known as newline character n
is appended to all user input from a terminal as it's telling the stream to start a new line. You can safely account for this by using std::getline
when checking for multiple lines of user input. The default behavior of std::getline
will read everything up to and including the newline character n
from the input stream object which is std::cin
in this case.
#include <iostream>
#include <string>
int main()
{
std::string name;
std::string state;
if (std::getline(std::cin, name) && std::getline(std::cin, state))
{
std::cout << "Your name is " << name << " and you live in " << state;
}
return 0;
}
Input:
"John"
"New Hampshire"
Output:
"Your name is John and you live in New Hampshire"
edited Feb 21 at 1:19
answered Feb 21 at 0:46
Justin Randall
1,6612812
1,6612812
add a comment |
add a comment |
protected by StoryTeller Dec 6 '17 at 8:53
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
I believe
std::cin >> name && std::cin >> std::skipws && std::getline(std::cin, state)
should also work as expected. (In addition to the answers below).– jww
Nov 11 at 17:30