[InstCombine] Remove hasNoInfs check for pow(C,y) -> exp2(log2(C)*y)

[foad]

Summary

We already check hasNoNaNs and that x is finite and strictly positive.
That only leaves the following special cases (taken from the Linux man
page for pow):

If the absolute value of x is less than 1, and y is negative infinity, the result is positive infinity.
If the absolute value of x is greater than 1, and y is negative infinity, the result is +0.
If the absolute value of x is less than 1, and y is positive infinity, the result is +0.
If the absolute value of x is greater than 1, and y is positive infinity, the result is positive infinity.

The transformation preserves all of these, so there is no need to limit
it to hasNoInfs.

[evandro]

pow(∞, y) → 0 if y < 0 or ∞ if y > 0
exp2(log2(∞) * y) → exp2(∞ * y) → exp2(∞) → 0 if y < 0 or ∞ if y > 0

[foad]

In D79409#2023511, @evandro wrote:

pow(∞, y) → 0 if y < 0 or ∞ if y > 0
exp2(log2(∞) * y) → exp2(∞ * y) → exp2(∞) → 0 if y < 0 or ∞ if y > 0

We check that x is a finite constant before doing the transformation. It's only y that could be infinite.

[evandro]

pow(x, -∞) → ∞ if |x| < 1 or 0 if |x| > 1
pow(x, +∞) → 0 if |x| < 1 or ∞ if |x| > 1
exp2(log2(x) * -∞) → exp2(±∞) → ∞ if x < 2 or 0 if x > 2
exp2(log2(x) * +∞) → exp2(±∞) → 0 if x < 2 or ∞ if x > 2

[foad]

In D79409#2023671, @evandro wrote:

exp2(log2(x) * -∞) → exp2(±∞) → ∞ if x < 2 or 0 if x > 2
exp2(log2(x) * +∞) → exp2(±∞) → 0 if x < 2 or ∞ if x > 2

No, the sign of log2(x) depends on whether x is less or greater than 1, not less or greater than 2.

[evandro]

In D79409#2023674, @foad wrote:

No, the sign of log2(x) depends on whether x is less or greater than 1, not less or greater than 2.

You're right.

pow(x, -∞) → ∞ if x < 1 or 0 if x > 1
pow(x, +∞) → 0 if x < 1 or ∞ if x > 1

exp2(log2(x) * -∞) → exp2(±∞) → ∞ if x < 1 or 0 if x > 1
exp2(log2(x) * +∞) → exp2(±∞) → 0 if x < 1 or ∞ if x > 1

Therefore the identity holds.

[foad]

Thanks for the careful checking. Actually there is one problematic case:
pow(1, ±∞) → 1
exp2(log2(1) * ±∞) → exp2(0 * ±∞) → exp2(NaN) → NaN
I guess we never hit this case in practice because it will have been simplified elsewhere, but I will add something to the patch to handle it.

[foad]

Rebase. Check for x == 1.

[spatel]

llvm/lib/Transforms/Utils/SimplifyLibCalls.cpp

1691–1692

There's no way for Base to be 1.0 after here, so we can use an assert rather than 'if' clause in the new code.

[foad]

Assert.

[foad]

Fix silly mistake in last revision.

[evandro]

llvm/lib/Transforms/Utils/SimplifyLibCalls.cpp

1570

This comment makes more sense immediately before the assert(). Personally, it could also be shorter.

[foad]

Move and shorten the comment.

[foad]

Ping!

[spatel]

LGTM