Pointers ignoring unsigned trait in C?

Question

I'm curious why the following code returns negative numbers even though every single variable and fuction initialization in this program is an unsigned int. I thought they always were supposed to be positive?.

#include <stdio.h>

unsigned int main ()
{
unsigned int ch = 0;
unsigned int asteriskValue = 0;
while(ch!=27)
{
    ch = getch();
        if (ch == 224)
        {
            increaseDecrease(&asteriskValue);
        }
    printf("%d",asteriskValue);
}
}

void increaseDecrease(unsigned int *value)
{
unsigned int upOrDown;
upOrDown = getch ();
if(upOrDown == 72)
    (*value)--;
else if(upOrDown == 80)
    (*value)++;
}

unsigned integers are not only positive, but can also be 0. Using The wrong format-specifier invokes undefined behaviour. — too honest for this site
unsigned is not a trait, but a type specifier. C does not support traits. — too honest for this site

Alexandre Mazel Alexandre Mazel · Accepted Answer · 2015-09-29T22:54:20

I think your results is really a positive value, but printing it with a "%d" flag, tells printf to interpret it as a signed value, as if you were casting on the fly the type: (signed)asteriskValue

Try printing it with a "%u" formatting instead of "%d".

Pointers ignoring unsigned trait in C?

1 Answers