Why does C interpret unsigned short int as unsigned long?

Question

In this main function, I get an error message that I should not use %hu (format specifier for unsigned short), and that the type of unsigned short appears to be unsigned long. How could that be the case?

#include <stdio.h>

int main (void) {
    unsigned short int a, b, c;
    printf("sizeof short int - %hu bytes \n\n", sizeof(unsigned short int));
}

The shell output is:

warning: format specifies type
      'unsigned short' but the argument has type
      'unsigned long' [-Wformat]
  ...%hu bytes \n\n", sizeof(unsigned short int));
     ~~~              ^~~~~~~~~~~~~~~~~~~~~~~~~~
     %lu

score 5 · Answer 1 · answered Aug 23 '20 at 12:55

Result of sizeof operator is size_t regardless of what is passed to that.

Quote from N1570 6.5.3.4 The sizeof and _Alignof operators:

2 The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

5 The value of the result of both operators is implementation-defined, and its type (an unsigned integer type) is size_t, defined in <stddef.h> (and other headers).

The format specifier to print size_t is %zu.

Why does C interpret unsigned short int as unsigned long?

1 Answers1