Bitwise Less than or Equal to

Question

There seems to be some kind of misconception that this is for a contest. I\'m trying to work through an assignment and I\'ve been stuck on this for an hour now.

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Answer 1

Here is my implementation(spend around 3 hours...)

int
isLessOrEqual(int x, int y)
{
    int a = y + ~x + 1;
    int b = a & 1 << 31 & a; // !b => y >= x, but maybe overflow
    int c = !!(x & (1 << 31)) & !(y & (1 << 31)); // y > 0, x < 0
    int d = !(x & (1 << 31)) & !!(y & (1 << 31)); // x > 0, y < 0

    int mask1 = !c + ~0;

    // if y > 0 && x < 0, return 1. else return !b
    int ans = ~mask1 & !b | mask1 & 1;

    int mask2 = !d + ~0;

  // if y < 0 && x > 0, return 0, else return ans
    return ~mask2 & ans | mask2 & 0;
}

• y - x == y + ~x + 1

• a & 1 << 31 & a is simplify from !(!(a & (1 << 31)) | !a)

The logic is:

if `y > 0 && x < 0`
    return true  
if `x > 0 && y < 0`
    return false

return y >= x

Why not just y >= x directly? because overflow may happen. So I have to early return to avoid overflow.