This commit suppresses turning loops like this into "(bitwidth - ctlz(input))".
unsigned foo(unsigned input) {
unsigned num = 0;
do {
++num;
input >>= 1;
} while (input != 0);
return num;
}
The loop version returns a value of 1 for both an input of 0 and an input of 1. Converting to a naive ctlz does not preserve that.
Theoretically we could do better if we checked isKnownNonZero or we could insert a select to handle the divergence. But until we have motivating cases for that, this is the easiest solution.
llvm-svn: 336864
This loop executes one iteration without checking the input value. This produces a count of 1 for an input of 0 and 1. We are turning this into 32 - ctlz(n), but that returns 0 if n is 0.
llvm-svn: 336862
In the 'detectCTLZIdiom' function support for loops that use LSHR instruction instead of ASHR has been added.
This supports creating ctlz from the following code.
int lzcnt(int x) {
int count = 0;
while (x > 0) {
count++;
x = x >> 1;
}
return count;
}
Patch by Olga Moldovanova
Differential Revision: https://reviews.llvm.org/D48354
llvm-svn: 336509
When checking a select to see if it matches an abs, allow the true/false values
to be a sign-extension of the comparison value instead of requiring that they're
directly the comparison value, as all the comparison cares about is the sign of
the value.
This fixes a regression due to r333702, where we were no longer generating ctlz
due to isKnownNonNegative failing to match such a pattern.
Differential Revision: https://reviews.llvm.org/D47631
llvm-svn: 333927
Summary:
Loop idiom recognize tries to convert loops like
```
int foo(int x) {
int cnt = 0;
while (x) {
x >>= 1;
++cnt;
}
return cnt;
}
```
into calls to ctlz, but if x is initially negative this loop should be infinite.
It happens that the cases that motivated this change have an absolute value of x before the loop. So this patch restricts the transform to cases where we know x is positive. Note: We are relying on the absolute value of INT_MIN to be undefined so we can assume that the result is always positive.
Fixes PR37479
Reviewers: spatel, hfinkel, efriedma, javed.absar
Reviewed By: efriedma
Subscribers: dmgreen, llvm-commits
Differential Revision: https://reviews.llvm.org/D47348
llvm-svn: 333702
We currently recognize this idiom where x is signed and thus the shift in an ashr.
int cnt = 0;
while (x) {
x >>= 1; // arithmetic shift right
++cnt;
}
and turn it into (bitwidth - ctlz(x)). And if there is anything else in the loop we will create a new loop that runs that many times.
If x is initially negative, the shift result will never be 0 and thus the loop is infinite. If you put something with side effects in the loop, that side effect will now only happen bitwidth times instead of an infinite number of times.
So this transform is only safe for logical shift right (which we don't currently recognize) or if we can prove that x cannot be negative before the loop.
llvm-svn: 331493
Summary:
The following loops should be recognized:
i = 0;
while (n) {
n = n >> 1;
i++;
body();
}
use(i);
And replaced with builtin_ctlz(n) if body() is empty or
for CPUs that have CTLZ instruction converted to countable:
for (j = 0; j < builtin_ctlz(n); j++) {
n = n >> 1;
i++;
body();
}
use(builtin_ctlz(n));
Reviewers: rengolin, joerg
Differential Revision: http://reviews.llvm.org/D32605
From: Evgeny Stupachenko <evstupac@gmail.com>
llvm-svn: 303102
Craig Topper