Question

The language I use is C.

I have code as follows:

int sign_x=~(x>>31)+1;
int sign_y=~(y>>31)+1;
int sign=sign_x^sign_y;
return ((!sign)&(!(!(0x80000000&(y+(~x+1)))))+(sign&(!sign_x)));

When I set input x=2147483647[0x7fffffff],y=-2147483648[0x80000000],the result is 0.
I wonder why the result is 0 rather than 1 because the first part is 0 and the second part is 1.When I changed operator “+” to “|”,the result is OK.

Can anybody help me? Thanks

Answer 1

0 + 1 is 1, which means that the operator precedence is different than what you thought it is.

gcc is clever enough to emit a warning for this:

warning: suggest parentheses around ‘+’ in operand of ‘&’

(!sign) & (!(!(0x80000000&(y+(~x+1)))))+(sign&(!sign_x)));

add extra parentheses:

((!sign)&(!(!(0x80000000&(y+(~x+1)))))) + ((sign&(!sign_x))));

or, since the expression is quite complex, temporarily store parts of the result:

int part1 = ...
int part2 = ...
return part1 + part2;