Question

I could not identify how the following program outputs 6 and -250.

#include<stdio.h>

int main()
{
    unsigned char p=-250;
    printf("%d",p);
    unsigned int p1=-250;
    printf("%d",p1);
    return 0;
}

Being an unsigned integer it has to output only the positive values.How does the p value outputs 6? Please help me understand.

Answer 1

printf is not typesafe. It prints whatever you ask it to, and %d says “signed integer”. It is your responsibility to provide a varibale of matching type. Since the unsigned char is only 8 bits wide, the literal -250 wraps around to +6, which remains +6 when interpreted as a signed integer. Note that char and short int (and their signed/unsigned counterparts) all get promoted to int-types when passed via variadic arguments.