Why are C character literals ints instead of chars?

Question

In C++, sizeof(\'a\') == sizeof(char) == 1. This makes intuitive sense, since \'a\' is a character literal, and sizeof(char) == 1 as d

Answer 1

The original question is "why?"

The reason is that the definition of a literal character has evolved and changed, while trying to remain backwards compatible with existing code.

In the dark days of early C there were no types at all. By the time I first learnt to program in C, types had been introduced, but functions didn't have prototypes to tell the caller what the argument types were. Instead it was standardised that everything passed as a parameter would either be the size of an int (this included all pointers) or it would be a double.

This meant that when you were writing the function, all the parameters that weren't double were stored on the stack as ints, no matter how you declared them, and the compiler put code in the function to handle this for you.

This made things somewhat inconsistent, so when K&R wrote their famous book, they put in the rule that a character literal would always be promoted to an int in any expression, not just a function parameter.

When the ANSI committee first standardised C, they changed this rule so that a character literal would simply be an int, since this seemed a simpler way of achieving the same thing.

When C++ was being designed, all functions were required to have full prototypes (this is still not required in C, although it is universally accepted as good practice). Because of this, it was decided that a character literal could be stored in a char. The advantage of this in C++ is that a function with a char parameter and a function with an int parameter have different signatures. This advantage is not the case in C.

This is why they are different. Evolution...