Thread: 2 questions about bit fields and variable handling

1 question:

Code:

#include <stdio.h>

struct foo{
       int a:4;
       };

int main(void){
    struct foo s;
    
    s.a=10;
    
    printf("%d\n",s.a);
    getchar();
    
    return 0;
    
}

10 in binary is 1010, but this gives me output -6, shouldn't it be -2?

the forth digit of 10 in binary is 1 that means the result will be <0, the remaining part of this is 010 which is 2, so why does it give me -6 instead of -2?

2 question:

Code:

int x=0;
int y=++x; // y=1 and x=1

int x=0;
int y=x++; //y=0 and x=1

but if we have

Code:

y=0;
x=y==++y;

why is x here 1??? I mean 0!=1 right? I can't get it

also

Code:

x=y==y++;

here x is 0?? Why?

thanks in advance

1 answer:
hmmm...

2 answer:

Code:

y=0;
x=y==++y;

What happens is this
1) y += 1
2) y == y which is 1 == 1, true
3) x = true (the compiler gives true == 1)

Code:

x=y==y++;

I would guess you would get 1 again. But maybe you compiler does something like
1) a = y, b = y++
2) a == b
3) a = y = 0, b = y++ = 1. Thus a != b
In the previous one it would do the same, but
1) y+= 1
2) a = b
3) a = y = 1, b = y = 1. Thus a == b
since ++y will happen first

The problem with the second is that it could also do
3) b = y++ = 1, a = y = 1. Thus a == b
so it is ambiguous
It could also do
1) a = y, b = y
2) y += 1
3) a == b, true
or switch 2) with 3) in which case again true

The reason of a and b is because it needs temporarily variables for the result of the left and right side. The general idea is that you have something like foo() == boo() so temporarily variables are need for results.

These are ofcourse assumptions. You realise that in a real program you should never use code like that no matter what.

thanks a lot