Why is the size of the data type different when the value is directly passed to the sizeof operator?

Constants, like variables, have a type of their own:

• 6.5 : A floating point constant of type double
• 90000 : An integer constant of type int (if int is 32 bits) or long (if int is 16 bits)
• 'A' : A character constant of type int in C and char in C++

The sizes that are printed are the sizes of the above types.

Also, the result of the sizeof operator has type size_t. So when printing the proper format specifier to use is %zu, not %d.

Character constants in C (opposite to C++) have the type int. So this call

printf("%d",sizeof('A'));

outputs 4. That is sizeof( 'A' ) is equal to sizeof( int ).

From the C Standard (6.4.4.4 Character constants)

10 An integer character constant has type int....

On the other hand (6.5.3.4 The sizeof and alignof operators)

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

So the operand of the sizeof operator in this expression sizeof( 'A' ) has the type int while in this expression sizeof( a ) where a is declared like

char a = 'A';

the operand has the type char.

Pay attention to that calls like this

printf("%d",sizeof(6.5));

use incorrect conversion format specifier. You have to write

printf("%zu",sizeof(6.5));

Also in the above call there is used a constant of the type double while in this call

printf("%zu",sizeof(c));

the variable c has the type float.

You could get the same result for these calls if the first call used a constant of the type float like

printf("%zu",sizeof(6.5f));