In C, why is sizeof(char) 1, when 'a' is an int?

Question

I tried

printf("%d, %d\\n", sizeof(char), sizeof('c'));

and got 1, 4 as output. If size of a character is one, why does 'c' give me 4? I guess it's because it's an integer. So when I do char ch = 'c'; is there an implicit conversion happening, under the hood, from that 4 byte value to a 1 byte value when it's assigned to the char variable?

Answer 1

In C 'a' is an integer constant (!?!), so 4 is correct for your architecture. It is implicitly converted to char for the assignment. sizeof(char) is always 1 by definition. The standard doesn't say what units 1 is, but it is often bytes.

Answer 2

Th C standard says that a character literal like 'a' is of type int, not type char. It therefore has (on your platform) sizeof == 4. See this question for a fuller discussion.

Answer 3

It is the normal behavior of the sizeof operator (See Wikipedia ):

• For a datatype, sizeof returns the size of the datatype. For char , you get 1.
• For an expression, sizeof returns the size of the type of the variable or expression. As a character literal is typed as int , you get 4.

Answer 4

This is covered in ISO C11 6.4.4.4 Character constants though it's largely unchanged from earlier standards. That states, in paragraph /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.

Answer 5

According to the ANSI C standards, a char gets promoted to an int in the context where integers are used, you used a integer format specifier in the printf hence the different values. A char is usually 1 byte but that is implementation defined based on the runtime and compiler.