CODEWITHSUNDEEP
c++cgcc
Pratt
Pratt

Reputation: 861

Single, double quotes and sizeof('a') in C/C++

I was looking at the question Single quotes vs. double quotes in C or C++. I couldn't completely understand the explanation given so I wrote a program:

#include <stdio.h>
int main()
{
  char ch = 'a';
  printf("sizeof(ch) :%d\n", sizeof(ch));
  printf("sizeof(\'a\') :%d\n", sizeof('a'));
  printf("sizeof(\"a\") :%d\n", sizeof("a"));
  printf("sizeof(char) :%d\n", sizeof(char));
  printf("sizeof(int) :%d\n", sizeof(int));
  return 0;
}

I compiled them using both gcc and g++ and these are my outputs:

gcc:

sizeof(ch)   : 1  
sizeof('a')  : 4  
sizeof("a")  : 2  
sizeof(char) : 1  
sizeof(int)  : 4

g++:

sizeof(ch)   : 1  
sizeof('a')  : 1  
sizeof("a")  : 2  
sizeof(char) : 1  
sizeof(int)  : 4

The g++ output makes sense to me and I don't have any doubt regarding that. In gcc, what is the need to have sizeof('a') to be different from sizeof(char)? Is there some actual reason behind it or is it just historical?

Also in C if char and 'a' have different size, does that mean that when we write char ch = 'a';, we are doing implicit type-conversion?

Upvotes: 58

Views: 6580

Answers (2)

ucefkh
ucefkh

Reputation: 2531

because there is no char just intgers linked int a character

like a is 62 i guess

if you try printf("%c",62); you will see a character

Upvotes: 1

Daniel Fischer
Daniel Fischer

Reputation: 183878

In C, character constants such as 'a' have type int, in C++ it's char.

Regarding the last question, yes,

char ch = 'a';

causes an implicit conversion of the int to char.

Upvotes: 61

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