Mixing
Does char == char implicitly cast both operands to int?
I am freshening up my C++ skills, using a copy of the german book "C++ - Das Übungsbuch"; 5. Auflage from mitp.
Exercise 8.1 poses the following question (translated and reduced to the relevant content):
8.1 Given the following definition:
char c = 'q';
Determine the type to which the operands will be converted during evaluation of the following statements:
a)
c == 'Q'
My answer would have been char, as c is declared as a char and as far as I understood, 'Q' is a character literal. The solution however states:
a) The type of both operands will be converted to
int
This confuses me. Is it really the case, that a char == char comparison is implicitly converted to int == int and why would this be done? If the question were
a)
c == 82
I could have understood that 82 would be interpreted as an integer constant and therefor c would be converted to int as well, but the answer explicitly mentions both operands to be converted.
Can someone explain this to me please?
c++
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
add a comment |
I am freshening up my C++ skills, using a copy of the german book "C++ - Das Übungsbuch"; 5. Auflage from mitp.
Exercise 8.1 poses the following question (translated and reduced to the relevant content):
8.1 Given the following definition:
char c = 'q';
Determine the type to which the operands will be converted during evaluation of the following statements:
a)
c == 'Q'
My answer would have been char, as c is declared as a char and as far as I understood, 'Q' is a character literal. The solution however states:
a) The type of both operands will be converted to
int
This confuses me. Is it really the case, that a char == char comparison is implicitly converted to int == int and why would this be done? If the question were
a)
c == 82
I could have understood that 82 would be interpreted as an integer constant and therefor c would be converted to int as well, but the answer explicitly mentions both operands to be converted.
Can someone explain this to me please?
c++
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
5
Yes, technically both operands get casted toint. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.
– Sam Varshavchik
Jan 1 at 16:35
2
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50
add a comment |
I am freshening up my C++ skills, using a copy of the german book "C++ - Das Übungsbuch"; 5. Auflage from mitp.
Exercise 8.1 poses the following question (translated and reduced to the relevant content):
8.1 Given the following definition:
char c = 'q';
Determine the type to which the operands will be converted during evaluation of the following statements:
a)
c == 'Q'
My answer would have been char, as c is declared as a char and as far as I understood, 'Q' is a character literal. The solution however states:
a) The type of both operands will be converted to
int
This confuses me. Is it really the case, that a char == char comparison is implicitly converted to int == int and why would this be done? If the question were
a)
c == 82
I could have understood that 82 would be interpreted as an integer constant and therefor c would be converted to int as well, but the answer explicitly mentions both operands to be converted.
Can someone explain this to me please?
c++
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
I am freshening up my C++ skills, using a copy of the german book "C++ - Das Übungsbuch"; 5. Auflage from mitp.
Exercise 8.1 poses the following question (translated and reduced to the relevant content):
8.1 Given the following definition:
char c = 'q';
Determine the type to which the operands will be converted during evaluation of the following statements:
a)
c == 'Q'
My answer would have been char, as c is declared as a char and as far as I understood, 'Q' is a character literal. The solution however states:
a) The type of both operands will be converted to
int
This confuses me. Is it really the case, that a char == char comparison is implicitly converted to int == int and why would this be done? If the question were
a)
c == 82
I could have understood that 82 would be interpreted as an integer constant and therefor c would be converted to int as well, but the answer explicitly mentions both operands to be converted.
Can someone explain this to me please?
c++
c++
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
edited Jan 1 at 16:51
T3 H40
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
asked Jan 1 at 16:31
T3 H40T3 H40
1,48072330
1,48072330
5
Yes, technically both operands get casted toint. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.
– Sam Varshavchik
Jan 1 at 16:35
2
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50
add a comment |
5
Yes, technically both operands get casted toint. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.
– Sam Varshavchik
Jan 1 at 16:35
2
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50
5
5
Yes, technically both operands get casted to
int. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.– Sam Varshavchik
Jan 1 at 16:35
Yes, technically both operands get casted to
int. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.– Sam Varshavchik
Jan 1 at 16:35
2
2
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50
add a comment |
2 Answers
2
active
oldest
votes
The book is talking about usual arithmetic conversion which happens if both operands have arithmetic types (which is true for all integer types, including char). That leads to integral promotion.
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
add a comment |
I think, that it uses for the comparison the ascii number corresponding to your character.
For example it converts q to 113 and Q to 81.
So it should compare:
'Q'=='q' //81==113
They are casted into integer because C interprets character as number of the ascii table.
See https://www.rapidtables.com/code/text/ascii-table.html
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
I don't think this question is as much about internalcharrepresentation as it is about how integral comparisons work according to the standard.
– Moira
Jan 1 at 16:46
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The book is talking about usual arithmetic conversion which happens if both operands have arithmetic types (which is true for all integer types, including char). That leads to integral promotion.
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
add a comment |
The book is talking about usual arithmetic conversion which happens if both operands have arithmetic types (which is true for all integer types, including char). That leads to integral promotion.
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
add a comment |
The book is talking about usual arithmetic conversion which happens if both operands have arithmetic types (which is true for all integer types, including char). That leads to integral promotion.
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
The book is talking about usual arithmetic conversion which happens if both operands have arithmetic types (which is true for all integer types, including char). That leads to integral promotion.
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
answered Jan 1 at 16:40
Some programmer dudeSome programmer dude
302k25264424
302k25264424
add a comment |
add a comment |
I think, that it uses for the comparison the ascii number corresponding to your character.
For example it converts q to 113 and Q to 81.
So it should compare:
'Q'=='q' //81==113
They are casted into integer because C interprets character as number of the ascii table.
See https://www.rapidtables.com/code/text/ascii-table.html
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
I don't think this question is as much about internalcharrepresentation as it is about how integral comparisons work according to the standard.
– Moira
Jan 1 at 16:46
add a comment |
I think, that it uses for the comparison the ascii number corresponding to your character.
For example it converts q to 113 and Q to 81.
So it should compare:
'Q'=='q' //81==113
They are casted into integer because C interprets character as number of the ascii table.
See https://www.rapidtables.com/code/text/ascii-table.html
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
I don't think this question is as much about internalcharrepresentation as it is about how integral comparisons work according to the standard.
– Moira
Jan 1 at 16:46
add a comment |
I think, that it uses for the comparison the ascii number corresponding to your character.
For example it converts q to 113 and Q to 81.
So it should compare:
'Q'=='q' //81==113
They are casted into integer because C interprets character as number of the ascii table.
See https://www.rapidtables.com/code/text/ascii-table.html
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
I think, that it uses for the comparison the ascii number corresponding to your character.
For example it converts q to 113 and Q to 81.
So it should compare:
'Q'=='q' //81==113
They are casted into integer because C interprets character as number of the ascii table.
See https://www.rapidtables.com/code/text/ascii-table.html
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
answered Jan 1 at 16:39
Massimo SimoneMassimo Simone
36
36
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
I don't think this question is as much about internalcharrepresentation as it is about how integral comparisons work according to the standard.
– Moira
Jan 1 at 16:46
add a comment |
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
I don't think this question is as much about internalcharrepresentation as it is about how integral comparisons work according to the standard.
– Moira
Jan 1 at 16:46
1
1
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
The question is about C++, and C and C++ behaves differently
– Basile Starynkevitch
Jan 1 at 16:43
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
Haven't read C++ tag sorry, my bad
– Massimo Simone
Jan 1 at 16:44
1
1
I don't think this question is as much about internal
char representation as it is about how integral comparisons work according to the standard.– Moira
Jan 1 at 16:46
I don't think this question is as much about internal
char representation as it is about how integral comparisons work according to the standard.– Moira
Jan 1 at 16:46
add a comment |
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5
Yes, technically both operands get casted to
int. However, most compilers will not generate the actual code to do that, because a compiler is permitted to perform any optimization that has no observable effect, and it's pretty much impossible to observe the effect of this optimization. But, technically, both are promoted to int.– Sam Varshavchik
Jan 1 at 16:35
2
Just a bit of terminology: this question is about implicit conversions, not casts. A cast is something you write in your source code to tell the compiler to do a conversion. There is no such thing as an implicit cast.
– Pete Becker
Jan 1 at 16:48
@PeteBecker Thank you, I'll fix that. I already thought that cast wouldn't be the correct term here, but I was lost for the correct word, which you now provided. :)
– T3 H40
Jan 1 at 16:50