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[LoopIdiomRecognize] Add a test case showing a loop we turn into ctlz that we shouldn't. 

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
This commit is contained in:
Craig Topper 2018-07-11 22:17:26 +00:00
parent 752ddbd068
commit ef08aec935
1 changed files with 36 additions and 0 deletions
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36
llvm/test/Transforms/LoopIdiom/X86/ctlz.ll
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@ -483,3 +483,39 @@ while.end: ; preds = %while.cond.while.en
ret i32 %cnt.0.lcssa
}
; FIXME: We should not transform this loop. It returns 1 for an input of both
; 0 and 1.
;
; int ctlz_bad(unsigned n)
; {
; int i = 0;
; do {
; i++;
; n >>= 1;
; } while(n != 0) {
; return i;
; }
;
; ALL: entry
; ALL-NEXT: %0 = call i32 @llvm.ctlz.i32(i32 %n, i1 false)
; ALL-NEXT: %1 = sub i32 32, %0
; ALL-NEXT: br label %while.cond
; ALL: %inc.lcssa = phi i32 [ %1, %while.cond ]
; ALL: ret i32 %inc.lcssa
; Function Attrs: norecurse nounwind readnone uwtable
define i32 @ctlz_bad(i32 %n) {
entry:
br label %while.cond
while.cond: ; preds = %while.cond, %entry
%n.addr.0 = phi i32 [ %n, %entry ], [ %shr, %while.cond ]
%i.0 = phi i32 [ 0, %entry ], [ %inc, %while.cond ]
%shr = lshr i32 %n.addr.0, 1
%tobool = icmp eq i32 %shr, 0
%inc = add nsw i32 %i.0, 1
br i1 %tobool, label %while.end, label %while.cond
while.end: ; preds = %while.cond
ret i32 %inc
}