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by tcpekin 1700 days ago
Why is this compiler optimization beneficial with -ffinite-math-only and -fno-signed-integers?

From

    if (x > y) {
      do_something();
    } else {
      do_something_else();
    }
to the form

    if (x <= y) {
      do_something_else();
    } else {
      do_something();
    }
What happens when x or y are NaN?
2 comments
progbits 1700 days ago
It is always false. In the first block do_something_else() gets executed when either is NaN, in the second it is do_something().
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zro 1700 days ago
> NaN is unordered: it is not equal to, greater than, or less than anything, including itself. x == x is false if the value of x is NaN [0]

My read of this is that comparisons involving NaN on either side always evaluate to false.

In the first one if X or Y is NaN then you'll get do_something_else, and in the second one you'll get do_something.

As far as why one order would be more optimal than the other, I'm not sure. Maybe something to do with branch prediction?

[0] https://www.gnu.org/software/libc/manual/html_node/Infinity-...

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