I'm confused with a output . So I'm expect

问题:

   #include <stdio.h>
    {
        char num = 127;
        num = num + 1;
        printf("%d", num);
      return 0;
     }

Output is : -128 should I add output shot

回答1:

char is a that, on most systems, takes 1 byte (8 bits). Your implementation seems to have char represent a signed type, however on other implementations it could be unsigned. The maximum value for a signed type is 2^(n-1)-1, where n is the number of bits. So the maximum value of char is 2^(8-1)-1=2^7-1=128-1=127. The minimum value is actually -2^(n-1). This means the minimun value is -128. When you add something that goes over the maximum value, it overflows and loops back to the minimum value. Hence, 127+1=-128 if you are doing char arithmetic.

You never use char for arithmetic. Use signed char or unsigned char instead. If you replace your char with unsigned char the program would print 128 as expected. Just note that the overflow can still happen (unsigned types have a range from 0 to 2^n-1, so unsigned char overflows if you add 1 to 255, giving you 0).

回答2:

char is a signed byte, that's why you are seeing that value.

as others mentioned you should read Range of signed char

but to quickly see how it works you can take a look at this example,

#include <stdio.h>
int main()
{
        char num;
        for(num=0; num<127;++num)
                printf("%d\n",num);
        printf("%d\n", num);
        num++;
        printf("%d\n", num);
        return 0;
}