size of char expression

Question

Can somebody throw a light on how sizeof(char expression) will be sizeof(int)?

int main()
{
char a, b;
printf("%d\n", sizeof(a+b));
return 0;
}

The program prints 4 as output. I was expecting it to be sizeof(char) i.e., 1

Answer 1

Integer promotion.

When small integral types (smaller than int, such as char, short, etc) are in arithmetic expression, they are automatically promoted to int.

---

C11 §6.3.1.1 Boolean, characters, and integers

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. ⁵⁸⁾

And in the footnote:

58) The integer promotions are applied only: as part of the usual arithmetic conversions, to certain argument expressions, to the operands of the unary +, -, and ~ operators, and to both operands of the shift operators, as specified by their respective subclauses.

Answer 2

a+b is an int expression. a and b would be char expressions.

When using most operators in C, operands of integral type but narrower than int are promoted to int.