memory allocation of the variable having type char

Question

i have the following code:

int main() { 
     char x='a';
     printf("integer = %ld\ncharacter  = %ld\nx = %ld\n", sizeof(int),sizeof('a'),sizeof(x);  
     return 0;  
}

and the output is:

integer = 4
character = 4
x = 1

my question is that in why the size of the char variable x is not equal to the size of the 'a' .

what actually brings the difference?

thanks for the help

This isn't your actual code, is it. This shouldn't even compile. Give us the real code. — Chris Eberle, Mar 03 '13 at 21:33

score 3 · Answer 1 · answered Mar 03 '13 at 21:32

3

Because in C single character constants (e.g., 'a' or '0') are of type int not char. This is different in C++ where they are of type char.

answered Mar 03 '13 at 21:32

ouah

1

+1 especially for mentioning this as a difference from C++. I always wonder about such subtle differences, now I've got one. – Mar 03 '13 at 21:35
1

@H2CO3 May be, they deliberately introduced such subtleties so as to rebut when people say "C/C++" ;) – P.P Mar 03 '13 at 21:42
@KingsIndian Correct! – Mar 03 '13 at 21:43

1 Answers1