logoalt Hacker News

kazinator • today at 6:39 PM • 1 reply • view on HN

The inline function receives the operands as arguments, and so whatever they are, they get converted to float. Thus the inline code is effectively like this:

  float v = (float) ((float) a) * ((float) b) + c;
Since v is float, the cast representing the return conversion can be omitted:

  float v = ((float) a) * ((float) b) + c;
Now, if a and b are already float, then it's equivalent. Otherwise not; if they are double or int, we get double or int multiplication in the original open code.

Replies

jcranmer • today at 7:03 PM

> Now, if a and b are already float, then it's equivalent.

Not necessarily! Floating-point contraction is allowable essentially within statements but not across them. By assigning the result of a * b into a value, you prohibit contraction from being able to contract with the addition into an FMA.

In practice, every compiler has fast-math flags which says stuff it and allows all of these optimizations to occur across statements and even across inline boundaries.

(Then there's also the issue of FLT_EVAL_METHOD, another area where what the standard says and what compilers actually do are fairly diametrically opposed.)

➕ show 2 replies