In java if “char c = 'a' ” why does “c = c + 1” not compile?
up vote
2
down vote
favorite
I tried to compile the following code:
public static void main(String args){
for (char c = 'a'; c <='z'; c = c + 1) {
System.out.println(c);
}
}
When I try to compile, it throws:
Error:(5, 41) java: incompatible types: possible lossy conversion from
int to char
The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.
I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.
java casting compiler-errors char primitive-types
edited Nov 12 at 11:01
Mickael
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
add a comment |
up vote
2
down vote
favorite
I tried to compile the following code:
public static void main(String args){
for (char c = 'a'; c <='z'; c = c + 1) {
System.out.println(c);
}
}
When I try to compile, it throws:
Error:(5, 41) java: incompatible types: possible lossy conversion from
int to char
The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.
I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.
java casting compiler-errors char primitive-types
edited Nov 12 at 11:01
Mickael
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I tried to compile the following code:
public static void main(String args){
for (char c = 'a'; c <='z'; c = c + 1) {
System.out.println(c);
}
}
When I try to compile, it throws:
Error:(5, 41) java: incompatible types: possible lossy conversion from
int to char
The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.
I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.
java casting compiler-errors char primitive-types
edited Nov 12 at 11:01
Mickael
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
I tried to compile the following code:
public static void main(String args){
for (char c = 'a'; c <='z'; c = c + 1) {
System.out.println(c);
}
}
When I try to compile, it throws:
Error:(5, 41) java: incompatible types: possible lossy conversion from
int to char
The thing is, it does work if I write c = (char)(c + 1), c += 1 or c++.
I checked and the compiler throws a similar error when I try char c = Character.MAX_VALUE + 1; but I see no way that the value of 'c' can pass 'char' type maximum in the original function.
java casting compiler-errors char primitive-types
java casting compiler-errors char primitive-types
edited Nov 12 at 11:01
Mickael
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
edited Nov 12 at 11:01
Mickael
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
edited Nov 12 at 11:01
Mickael
2,90521528
edited Nov 12 at 11:01
Mickael
2,90521528
edited Nov 12 at 11:01
Mickael
2,90521528
2,90521528
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
asked Nov 12 at 10:54
רעי וייס ליפשיץ
165
165
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
c + 1 is an int, as the operands undergo binary numeric promotion:
cis achar
1is anint
so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.
As for the things that "work":
c = (char)(c + 1)is explicitly casting the expression tochar, so its value is compatible with the variable's type;
c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.
c++is of typechar, so no cast is required.
answered Nov 12 at 10:56
Andy Turner
80k879133
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" soc + (char) 1wouldn't have worked.
– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why doeschar c = 97 + 1work? wouldn't java see97 + 1as an int?
– רעי וייס ליפשיץ
Nov 12 at 22:41
|
show 1 more comment
up vote
0
down vote
First you are declaring c as char than you are using it as an int
answered Nov 12 at 10:57
Brian
527
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
c + 1 is an int, as the operands undergo binary numeric promotion:
cis achar
1is anint
so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.
As for the things that "work":
c = (char)(c + 1)is explicitly casting the expression tochar, so its value is compatible with the variable's type;
c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.
c++is of typechar, so no cast is required.
answered Nov 12 at 10:56
Andy Turner
80k879133
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" soc + (char) 1wouldn't have worked.
– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why doeschar c = 97 + 1work? wouldn't java see97 + 1as an int?
– רעי וייס ליפשיץ
Nov 12 at 22:41
|
show 1 more comment
up vote
8
down vote
accepted
c + 1 is an int, as the operands undergo binary numeric promotion:
cis achar
1is anint
so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.
As for the things that "work":
c = (char)(c + 1)is explicitly casting the expression tochar, so its value is compatible with the variable's type;
c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.
c++is of typechar, so no cast is required.
answered Nov 12 at 10:56
Andy Turner
80k879133
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" soc + (char) 1wouldn't have worked.
– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why doeschar c = 97 + 1work? wouldn't java see97 + 1as an int?
– רעי וייס ליפשיץ
Nov 12 at 22:41
|
show 1 more comment
up vote
8
down vote
accepted
up vote
8
down vote
accepted
c + 1 is an int, as the operands undergo binary numeric promotion:
cis achar
1is anint
so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.
As for the things that "work":
c = (char)(c + 1)is explicitly casting the expression tochar, so its value is compatible with the variable's type;
c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.
c++is of typechar, so no cast is required.
answered Nov 12 at 10:56
Andy Turner
80k879133
c + 1 is an int, as the operands undergo binary numeric promotion:
cis achar
1is anint
so c has to be widened to int to make it compatible for addition; and the result of the expression is of type int.
As for the things that "work":
c = (char)(c + 1)is explicitly casting the expression tochar, so its value is compatible with the variable's type;
c += 1is equivalent toc = (char) ((c) + (1)), so it's basically the same as the previous one.
c++is of typechar, so no cast is required.
answered Nov 12 at 10:56
Andy Turner
80k879133
edited Nov 12 at 11:04
answered Nov 12 at 10:56
Andy Turner
80k879133
answered Nov 12 at 10:56
Andy Turner
80k879133
answered Nov 12 at 10:56
Andy Turner
80k879133
80k879133
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" soc + (char) 1wouldn't have worked.
– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why doeschar c = 97 + 1work? wouldn't java see97 + 1as an int?
– רעי וייס ליפשיץ
Nov 12 at 22:41
|
show 1 more comment
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" soc + (char) 1wouldn't have worked.
– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why doeschar c = 97 + 1work? wouldn't java see97 + 1as an int?
– רעי וייס ליפשיץ
Nov 12 at 22:41
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
Good answer, only suggestion would be to mention why you can set an int to a char, but not a char to an int (possible lossy conversion due to assigning a bigger data type to a smaller data type)
– Eamon Scullion
Nov 12 at 11:06
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
First of all, thank you for the good answer. Second, if I understand correctly, that mean java takes 'c + 1' and instead of casting '1' to a char it casts 'c' to an int in order to add them together. Is this because converting an int to char has a potential of losing data?
– רעי וייס ליפשיץ
Nov 12 at 11:23
1
1
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ yes. Note also that even if you add two chars together, they are still widened to int first. (Same is true of byte and short).
– Andy Turner
Nov 12 at 12:13
@רעיוייסליפשיץ "Note also that even if you add two chars together" so
c + (char) 1 wouldn't have worked.– Andy Turner
Nov 12 at 13:02
@רעיוייסליפשיץ "Note also that even if you add two chars together" so
c + (char) 1 wouldn't have worked.– Andy Turner
Nov 12 at 13:02
Thanks, but that leaves me with a question, why does
char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?– רעי וייס ליפשיץ
Nov 12 at 22:41
Thanks, but that leaves me with a question, why does
char c = 97 + 1 work? wouldn't java see 97 + 1 as an int?– רעי וייס ליפשיץ
Nov 12 at 22:41
|
show 1 more comment
up vote
0
down vote
First you are declaring c as char than you are using it as an int
answered Nov 12 at 10:57
Brian
527
add a comment |
up vote
0
down vote
First you are declaring c as char than you are using it as an int
answered Nov 12 at 10:57
Brian
527
add a comment |
up vote
0
down vote
up vote
0
down vote
First you are declaring c as char than you are using it as an int
answered Nov 12 at 10:57
Brian
527
First you are declaring c as char than you are using it as an int
answered Nov 12 at 10:57
Brian
527
answered Nov 12 at 10:57
Brian
527
answered Nov 12 at 10:57
Brian
527
answered Nov 12 at 10:57
Brian
527
527
add a comment |
add a comment |
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