Return zero for negative integers

A friend just throw some code similar to following C# code:

int i = ...;
return i < 0 ? 0 : i;

That made me think. There's any "different" way to return zero开发者_Python百科 for negative integers, or current positive value? More specifically I'm looking for bitwise operations, if possible.

BTW, I'm aware of Math.Max(0, i);

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What's wrong with Math.Max?

You can do the equivalent without a branch using bitwise operations:

r = x ^ ((x ^ y) & -(x < y)); // == max(x, y)

If you substitute zero, it collapses to:

r = (y & -(0 < y)); // == max(0, y)

(Source: this list of bitwise tricks.)

If branches were extremely expensive on your platform, that might be worthwhile in some inner loop, I suppose, but it's pretty obscure and not the kind of thing I'd like to come across outside of an extremely time-sensitive function.

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How about:

int i = ...;

return i & ~(i >> 31);

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The below will do the trick and the code reads so well it practically don't need a comment ;)

((((0x80000000 & i) >> 31)^1) * 0xFFFFFFFF) & i

then again

int temp = (0x80000000 & i); //get the most significant bit (1 for negative 0 for positive)
temp = (temp>>31)^1; //more it to the least significant and not it (we now have 0 for negative numbers and one for positive)

temp *= 0xFFFFFFFF; //get a lof of F's if it's positive or a lot of zeros if the number was negative

temp = temp & i; //and the F's/zeros with the original number

and voila zero for all negative number and all positive are left unchanged

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Short answer: No.

Bit operators do something very different, or rather are used for different problems.

If you know the size of your integers, you could test the highest (most significant) bit; if it's 1, the number is negative and you can act on that. But that would be a heck of a lot more work than the simple "<" test.

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Not bitwise but different:

return (i + Math.abs(i))/2

EDIT:

return (int)(i/2f + Math.abs(i)/2f)