Does std::optional change signature of the function?
I need to make an optional argument with a default value in my function. Currently the signature looks something like this:
void func(int a, std::optional<int> b = 10)
and the function behaves in the following way:
func(15, 5); // works
func(15); // works
The question is: If I remove the explicit initialization for the optional argument, like this:
void func(int a, std::optional<int> b)
Then It seems like the signature of the function changes
func(15, 5); // works
func(15); // fails
Which makes me very confused about the purpose of the std::optional in the first place. What is it good for if not for creating optional arguments?
c++ c++17 stdoptional
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
add a comment |
I need to make an optional argument with a default value in my function. Currently the signature looks something like this:
void func(int a, std::optional<int> b = 10)
and the function behaves in the following way:
func(15, 5); // works
func(15); // works
The question is: If I remove the explicit initialization for the optional argument, like this:
void func(int a, std::optional<int> b)
Then It seems like the signature of the function changes
func(15, 5); // works
func(15); // fails
Which makes me very confused about the purpose of the std::optional in the first place. What is it good for if not for creating optional arguments?
c++ c++17 stdoptional
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
1
std::optionalis a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.
– 0x499602D2
Nov 14 '18 at 0:38
It does not make much sense to have anoptionalwith a default value that is not empty as it would only confuse people.
– Phil1970
Nov 14 '18 at 2:51
add a comment |
I need to make an optional argument with a default value in my function. Currently the signature looks something like this:
void func(int a, std::optional<int> b = 10)
and the function behaves in the following way:
func(15, 5); // works
func(15); // works
The question is: If I remove the explicit initialization for the optional argument, like this:
void func(int a, std::optional<int> b)
Then It seems like the signature of the function changes
func(15, 5); // works
func(15); // fails
Which makes me very confused about the purpose of the std::optional in the first place. What is it good for if not for creating optional arguments?
c++ c++17 stdoptional
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
I need to make an optional argument with a default value in my function. Currently the signature looks something like this:
void func(int a, std::optional<int> b = 10)
and the function behaves in the following way:
func(15, 5); // works
func(15); // works
The question is: If I remove the explicit initialization for the optional argument, like this:
void func(int a, std::optional<int> b)
Then It seems like the signature of the function changes
func(15, 5); // works
func(15); // fails
Which makes me very confused about the purpose of the std::optional in the first place. What is it good for if not for creating optional arguments?
c++ c++17 stdoptional
c++ c++17 stdoptional
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
edited Nov 14 '18 at 1:18
songyuanyao
90.5k11171234
90.5k11171234
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
asked Nov 14 '18 at 0:34
nikolaevranikolaevra
696
696
1
std::optionalis a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.
– 0x499602D2
Nov 14 '18 at 0:38
It does not make much sense to have anoptionalwith a default value that is not empty as it would only confuse people.
– Phil1970
Nov 14 '18 at 2:51
add a comment |
1
std::optionalis a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.
– 0x499602D2
Nov 14 '18 at 0:38
It does not make much sense to have anoptionalwith a default value that is not empty as it would only confuse people.
– Phil1970
Nov 14 '18 at 2:51
1
1
std::optional is a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.– 0x499602D2
Nov 14 '18 at 0:38
std::optional is a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.– 0x499602D2
Nov 14 '18 at 0:38
It does not make much sense to have an
optional with a default value that is not empty as it would only confuse people.– Phil1970
Nov 14 '18 at 2:51
It does not make much sense to have an
optional with a default value that is not empty as it would only confuse people.– Phil1970
Nov 14 '18 at 2:51
add a comment |
3 Answers
3
active
oldest
votes
What is it good for if not for creating optional arguments?
std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.
The class template
std::optionalmanages an optional contained value, i.e. a value that may or may not be present.
You can used it like
void func(int a, std::optional<int> b = std::nullopt) {
if (b) {
// if b contains a value
...
} else {
...
}
}
then
func(15, 5); // b will contain a value (i.e. `5`)
func(15); // b doesn't contain a value
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
add a comment |
std::optional<int> is still a concrete type despite being "optional" so, unless you have a default value for it in your function specification, you need to supply one.
You seem to be conflating the two definitions of optional here:
- the concrete type allowing you to store an object or lack thereof; and
- the optionality (if that's even a real word) of function arguments.
They are not the same thing.
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
add a comment |
Another use: optional return values:
// throws if cannot parse
auto parse_int(const std::string& s) -> int;
// returns std::nullopt if it cannot parse
auto try_parse_int(const std::string& s) -> std::optional<int>
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
What is it good for if not for creating optional arguments?
std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.
The class template
std::optionalmanages an optional contained value, i.e. a value that may or may not be present.
You can used it like
void func(int a, std::optional<int> b = std::nullopt) {
if (b) {
// if b contains a value
...
} else {
...
}
}
then
func(15, 5); // b will contain a value (i.e. `5`)
func(15); // b doesn't contain a value
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
add a comment |
What is it good for if not for creating optional arguments?
std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.
The class template
std::optionalmanages an optional contained value, i.e. a value that may or may not be present.
You can used it like
void func(int a, std::optional<int> b = std::nullopt) {
if (b) {
// if b contains a value
...
} else {
...
}
}
then
func(15, 5); // b will contain a value (i.e. `5`)
func(15); // b doesn't contain a value
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
add a comment |
What is it good for if not for creating optional arguments?
std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.
The class template
std::optionalmanages an optional contained value, i.e. a value that may or may not be present.
You can used it like
void func(int a, std::optional<int> b = std::nullopt) {
if (b) {
// if b contains a value
...
} else {
...
}
}
then
func(15, 5); // b will contain a value (i.e. `5`)
func(15); // b doesn't contain a value
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
What is it good for if not for creating optional arguments?
std::optional is not supposed to be used for optional argument what you expect; which requires default argument as your 1st code sample showed, std::optional won't change the language syntax.
The class template
std::optionalmanages an optional contained value, i.e. a value that may or may not be present.
You can used it like
void func(int a, std::optional<int> b = std::nullopt) {
if (b) {
// if b contains a value
...
} else {
...
}
}
then
func(15, 5); // b will contain a value (i.e. `5`)
func(15); // b doesn't contain a value
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
edited Nov 14 '18 at 4:20
Nicol Bolas
285k33472645
285k33472645
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
answered Nov 14 '18 at 0:42
songyuanyaosongyuanyao
90.5k11171234
90.5k11171234
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
add a comment |
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
Thank you, makes perfect sense to me. Found it really hard to get to the bottom of this from just cpp docs, so had to ask here.
– nikolaevra
Nov 14 '18 at 1:15
add a comment |
std::optional<int> is still a concrete type despite being "optional" so, unless you have a default value for it in your function specification, you need to supply one.
You seem to be conflating the two definitions of optional here:
- the concrete type allowing you to store an object or lack thereof; and
- the optionality (if that's even a real word) of function arguments.
They are not the same thing.
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
add a comment |
std::optional<int> is still a concrete type despite being "optional" so, unless you have a default value for it in your function specification, you need to supply one.
You seem to be conflating the two definitions of optional here:
- the concrete type allowing you to store an object or lack thereof; and
- the optionality (if that's even a real word) of function arguments.
They are not the same thing.
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
add a comment |
std::optional<int> is still a concrete type despite being "optional" so, unless you have a default value for it in your function specification, you need to supply one.
You seem to be conflating the two definitions of optional here:
- the concrete type allowing you to store an object or lack thereof; and
- the optionality (if that's even a real word) of function arguments.
They are not the same thing.
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
std::optional<int> is still a concrete type despite being "optional" so, unless you have a default value for it in your function specification, you need to supply one.
You seem to be conflating the two definitions of optional here:
- the concrete type allowing you to store an object or lack thereof; and
- the optionality (if that's even a real word) of function arguments.
They are not the same thing.
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
answered Nov 14 '18 at 0:36
paxdiablopaxdiablo
632k17012451669
632k17012451669
add a comment |
add a comment |
Another use: optional return values:
// throws if cannot parse
auto parse_int(const std::string& s) -> int;
// returns std::nullopt if it cannot parse
auto try_parse_int(const std::string& s) -> std::optional<int>
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
add a comment |
Another use: optional return values:
// throws if cannot parse
auto parse_int(const std::string& s) -> int;
// returns std::nullopt if it cannot parse
auto try_parse_int(const std::string& s) -> std::optional<int>
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
add a comment |
Another use: optional return values:
// throws if cannot parse
auto parse_int(const std::string& s) -> int;
// returns std::nullopt if it cannot parse
auto try_parse_int(const std::string& s) -> std::optional<int>
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
Another use: optional return values:
// throws if cannot parse
auto parse_int(const std::string& s) -> int;
// returns std::nullopt if it cannot parse
auto try_parse_int(const std::string& s) -> std::optional<int>
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
answered Nov 14 '18 at 4:26
bolovbolov
31k670130
31k670130
add a comment |
add a comment |
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1
std::optionalis a class just like any other so it must be initialized. The only way you can omit an argument is when one has a default argument.– 0x499602D2
Nov 14 '18 at 0:38
It does not make much sense to have an
optionalwith a default value that is not empty as it would only confuse people.– Phil1970
Nov 14 '18 at 2:51