How does unsigned short int differs from unsigned int

Question

#include<stdio.h>
 int main()
 {
    unsigned  short int  x=-10;
    short int y=-10;
    unsigned int z=-10000000;
    int m=-10000000;

    printf("x=%d y=%d z=%d m=%d",x,y,z,m);
    return 0;
 }

output=x=65526 y=-10 z=-10000000 m=-10000000

My query is How unsigned short intdiffer from unsigned int in the scenario of data holding. i.e x=65526 where as z=-10000000 why x is not equal -10 where as z can hold any data As short is 2 byte so twos complement -10 is 65526 but why not same in case of z

http://stackoverflow.com/questions/3812022/what-is-a-difference-between-unsigned-int-and-signed-int-in-c — Rulisp, Oct 12 '13 at 15:27

chux - Reinstate Monica · Answer 1 · 2013-10-12T15:59:08.547

When unsigned short int x=-10; occurs, x, being unsigned, gets a modulo or "wrapped" value of 65526 (OP sizeof(short) is 2). Something like x = power(256,sizeof(unsigned short)) - 10.

When x is printed, it is passed to printf() as an int (variadic promotion). OP's sizeof(int) is 4, so the 65526 fits in int. Then printf() sees a %d and prints "65526".

z Has a similar story, but sizeof(z) is 4. and gets initialized z = power(256,sizeof(unsigned)) - 10.

Your printf() is using the %d specifier for unsigned. OP should use %u.

printf("x=%u y=%d z=%u m=%d",x,y,z,m);

unsigned short int is guaranteed to cover at least the range 0 to 65535.

unsigned int is guaranteed to cover at least the range of unsigned short int. It may cover a wider range.

unsigned int is usually the native size the processor best uses - typically fastest.

As unsigned short int is, in some implementations, smaller than unsigned int. It is preferred for space saving with large arrays.

How does unsigned short int differs from unsigned int

1 Answers1