What is the format specifier for unsigned short int?

Question

I have the following program

#include <stdio.h>

int main(void)
{
    unsigned short int length = 10; 

    printf("Enter length : ");
    scanf("%u", &length);

    printf("value is %u \n", length);

    return 0;
}

Which when compiled using gcc filename.c issued the following warning (in the scanf() line).

warning: format ‘%u’ expects argument of type ‘unsigned int *’, but argument 2 has type ‘short unsigned int *’ [-Wformat]

I then referred the C99 specification - 7.19.6 Formatted input/output functions and couldn't understand the correct format specifier when using the length modifiers (like short, long, etc) with unsigned for int data type.

Is %u the correct specifier unsigned short int? If so why am I getting the above mentioned warning?!

EDIT: Most of the time, I was trying %uh and it was still giving the warning.

Answer 1

Try using the "%h" modifier:

scanf("%hu", &length);
        ^

ISO/IEC 9899:201x - 7.21.6.1-7

Specifies that a following d , i , o , u , x , X , or n conversion specifier applies to an argument with type pointer to short or unsigned short.

Answer 2

For scanf, you need to use %hu since you're passing a pointer to an unsigned short. For printf, it's impossible to pass an unsigned short due to default promotions (it will be promoted to int or unsigned int depending on whether int has at least as many value bits as unsigned short or not) so %d or %u is fine. You're free to use %hu if you prefer, though.

Answer 3

From the Linux manual page:

h      A  following  integer conversion corresponds to a short int or unsigned short int argument, or a fol‐
       lowing n conversion corresponds to a pointer to a short int argument.

So to print an unsigned short integer, the format string should be "%hu".

Answer 4

Here is a good table for printf specifiers. So it should be %hu for unsigned short int.

And link to Wikipedia "C data types" too.

Answer 5

For printf, it's impossible to pass an unsigned short due to default promotions (it will be promoted to int or unsigned int depending on whether int has at least as many value bits as unsigned short or not) so %d or %u is fine.

But still he can manually convert value of this short int to a char array of '0' and '1' and then use another loop to create a short int value and print it I guess.

Let us do it this way:

struct byte3data {
    char a;
    char b;
    char c;
};
void my_printf(byte3data myOwnType) {
    // Writing 3 bytes with printf() in the way we want
}

printf is just basic tool, and this this how it should be treated I think. You can manage bits in any way you want. And with floating point arithmetic You can say yes or no but also how much. Try to change printf to printfs in your sentence and think about it again.