Question

Why a char variable gets ‘b’ from assignment of ‘ab’, rather ‘a’?

char c = 'ab';

printf("c: %c\n", c);

Prints:

c: b

Answer 1

This is implementation defined as earlier answers already say.

My gcc handles ‘ab’ as an int. The following code:

printf( "sizeof('ab') = %zu \n", sizeof('ab') );
printf( "'ab' = 0x%08x \n", 'ab' );
printf( "'abc' = 0x%08x \n", 'abc' );

prints:

sizeof('ab') = 4
'ab' = 0x00006162
'abc' = 0x00616263

In your code, the line:

char c = 'ab';

Can be considered as:

char c = (char)(0x00006162 & 0xFF);

So c gets the value of the last char of ‘ab’. In this case it is ‘b’ (0x62).