Question

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char **argv){
    char i = -128; 
    int j = i; 
    printf("%d %u\n", j, j); 
   return 0;
}

result is -128 4294967168

what I think is

i: 10000000 

and after the assignment operator, do the sign extension

j: 11111111 11111111 11111111 10000000 

what I want to ask is how printf("%d",j) know to print -128 just use

the last byte? How it works?

Thx!

Answer 1

what I want to ask is how printf(“%d”,j) know to print -128 just use the last byte?

It doesn’t. It is told to print a signed int, so it takes the appropriate number of bytes -typically 4 nowadays – from the stack and interprets that bit pattern as a signed int.

When you assign a negative char to an int variable, as in int j = i; here, what happens is not really sign-extension, but – since all values a char can represent are also representable as an int – a value-preserving conversion, the char i is converted to an int with the same value.

On two’s complement machines, which are by far the most common nowadays, and also in ones’ complement, that value-preserving conversion happens to coincide with sign-extension, but if the representaion is sign-and magnitude, the conversion would be different.

Since -128 isn’t representable as a signed eight-bit integer in ones’ complement or sign-and magnitude, let’s look at what happens to the bit-patterns when converting -127 to a 32-bit signed integer with the same kind of representation:

Two’s complement:

10000001 -> 11111111 11111111 11111111 10000001

Ones’ complement:

10000000 -> 11111111 11111111 11111111 10000000

Sign-and-magnitude:

11111111 -> 10000000 00000000 00000000 01111111