Question

         #include<stdio.h>

         int main()
         {

           char ch = 'A';

           printf("%d\n",'ag');

           printf("%d\n",'a');

           printf("%d, %d, %d, %d", sizeof(ch), sizeof('a'), sizeof('Ag'), sizeof(3.14f));

          return 0;
         }

I used to have many doubts on the output of this question while running on g++ and gcc.

But I have cleared almost all the doubts by referring these links:

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

2. Single quotes vs. double quotes in C or C++

I still need to understand one thing about the output of this question.

Can someone please explain the output of printf("%d\n",'ag'); mentioned above in the program. How is it actually stored in the memory?

The output for the program on the Linux/GCC platform is:

24935
97
1, 4, 4, 4

Answer 1

The type of a single-quoted literal is int. So the size is typically large enough for more than one character’s worth of bits. The exact way the characters are interpreted is, as far as I know, implementation-dependent.

In your case, you’re getting a little-endian ordering:

• The ASCII value for 'a' is 97 (0x61)
• The ASCII value for 'g' is 103 (0x67)

Your value is 24935 = 0x6167, so you’re getting the 'a' in the higher byte and the 'g' in the lower.