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

using gcc on my MacBook, I try:

#include 
#define test(A) do{printf(#A":\t%i\n",sizeof(A));}while(0)
int main(void){
  test('a');
  test("a");
  test("");
  test(char);
  test(short);
  test(int);
  test(long);
  test((char)0x0);
  test((short)0x0);
  test((int)0x0);
  test((long)0x0);
  return 0;
};

which when run gives:

'a':    4
"a":    2
"":     1
char:   1
short:  2
int:    4
long:   4
(char)0x0:      1
(short)0x0:     2
(int)0x0:       4
(long)0x0:      4

which suggests that a character is 8 bits, like you suspect, but a character literal is an int.