Printing short int using various format specifiers

Question

Please have a look at this code:

#include <stdio.h>
int main(void)
{
short s = -1;

printf("sizeof(short) = %lu\n", sizeof(short));
printf("sizeof(int) = %lu\n", sizeof(int));
printf("sizeof(long) = %lu\n", sizeof(long));
printf("s = %hd\n", s);
printf("s = %d\n", s);
printf("s = %ld\n", s);

return 0;
}  

It gave output as :

sizeof(short) = 2  
sizeof(int) = 4  
sizeof(long) = 8  
s = -1  
s = -1  
s = 4294967295  

In last line why s = 4294967295 instead of s = -1 as through this question I came to know that In C when variable gets promoted, its value remains constant.

Answer 1

s is being promoted to an int, which here is a 4 byte type. This is happening in all 3 cases. In the first two, an int is what printf() will expect, as the format specifier is for a type which would be passed as an int. However in the last case, you have given a format specifier which expects an 8-byte type.

This is invoking undefined behaviour.

In your case it appears to have read zeros in the upper bytes of the value, effectively zero-extending to 64-bits the value which was already sign-extended to 32-bits. However you can't depend on the results of doing this - it might be reading memory or a register that is not consistently initialised. Tomorrow it could be different.

The promotion of arguments is not dependent on the format string - you must always ensure that you pass the correct arguments for the format you have specified. So an int will not be promoted to a long. You need to convert it yourself.

A smart compiler ought to give you a warning about this.

Answer 2

The variable does not get promoted to long while calling printf("s = %ld\n", s);. The right way to promote it to long is printf("s = %ld\n", (long) s);.