Question

Check out this program

#include<stdio.h>

int main (){

char c='a';
printf("%d %d", sizeof(c),sizeof('a'));
}

the output is 1 4
I know when we write a statement char c=’a’;

then how does it happen that in space of 1 byte (char c) some thing of 4 bytes (ASCII code) is stored why is there no overflow etc.

Answer 1

First, per ANSI/IEC 9899:1999(E) §6.4.4.4:

 10. An integer character constant has type int. The value of an integer character constant
containing a single character that maps to a single-byte execution character is the
numerical value of the representation of the mapped character interpreted as an integer. […]

§6.5.3.4:

 2. The sizeof operator yields the size (in bytes) of its operand, which may be an
expression or the parenthesized name of a type. The size is determined from the type of
the operand. […]

 3. When applied to an operand that has type char, unsigned char, or signed char,
(or a qualified version thereof) the result is 1. […]

As you can see, since the type of a character constant is int, for sizeof('a') we get sizeof(int), which is 4 on your platform. However, for sizeof(c), we get the size of a char, which is defined to be 1.

So why can we assign 'a' to a char?

§6.5.16.1:

 2. In simple assignment (=), the value of the right operand is converted to the type of the assignment expression and replaces the value stored in the object designated by the left operand.

So, the int that is 'a' is implicitly converted to a char. There’s an example in there as well, showing explicitly that ints can be implicitly converted to char.