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Simplify a README.txt entry significantly to expose the core issue.

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llvm-svn: 123556
This commit is contained in:
Chandler Carruth 2011-01-16 01:40:23 +00:00
parent 44bcf63348
commit a3261fcca5
1 changed files with 24 additions and 26 deletions
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50
lib/Target/README.txt
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@ -2162,37 +2162,35 @@ Also surprising is that %conv isn't simplified to 0 in %....exit.thread.i.i.
clang -O3 -fno-exceptions currently compiles this code:
void f(int N) {
std::vector<int> v(N);
for (int k = 0; k < N; ++k)
v[k] = 0;
extern void sink(void*); sink(&v);
void f(char* a, int n) {
__builtin_memset(a, 0, n);
for (int i = 0; i < n; ++i)
a[i] = 0;
}
into almost the same as the previous note, but replace its final BB with:
into:
for.body.lr.ph: ; preds = %cond.true.i.i.i.i
%mul.i.i.i.i.i = shl i64 %conv, 2
%call3.i.i.i.i.i = call noalias i8* @_Znwm(i64 %mul.i.i.i.i.i) nounwind
%0 = bitcast i8* %call3.i.i.i.i.i to i32*
store i32* %0, i32** %v8.sub, align 8, !tbaa !0
%add.ptr.i.i.i = getelementptr inbounds i32* %0, i64 %conv
store i32* %add.ptr.i.i.i, i32** %tmp4.i.i.i.i.i, align 8, !tbaa !0
call void @llvm.memset.p0i8.i64(i8* %call3.i.i.i.i.i, i8 0, i64 %mul.i.i.i.i.i, i32 4, i1 false)
store i32* %add.ptr.i.i.i, i32** %tmp3.i.i.i.i.i, align 8, !tbaa !0
%tmp18 = add i32 %N, -1
%tmp19 = zext i32 %tmp18 to i64
%tmp20 = shl i64 %tmp19, 2
%tmp21 = add i64 %tmp20, 4
call void @llvm.memset.p0i8.i64(i8* %call3.i.i.i.i.i, i8 0, i64 %tmp21, i32 4, i1 false)
br label %for.end
define void @_Z1fPci(i8* nocapture %a, i32 %n) nounwind {
entry:
%conv = sext i32 %n to i64
tail call void @llvm.memset.p0i8.i64(i8* %a, i8 0, i64 %conv, i32 1, i1 false)
%cmp8 = icmp sgt i32 %n, 0
br i1 %cmp8, label %for.body.lr.ph, label %for.end
First off, why (((zext %N - 1) << 2) + 4) instead of the ((sext %N) << 2) done
previously? (or better yet, re-use that one?)
for.body.lr.ph: ; preds = %entry
%tmp10 = add i32 %n, -1
%tmp11 = zext i32 %tmp10 to i64
%tmp12 = add i64 %tmp11, 1
call void @llvm.memset.p0i8.i64(i8* %a, i8 0, i64 %tmp12, i32 1, i1 false)
ret void
Then, the really painful one is the second memset, of the same memory, to the
same value.
for.end: ; preds = %entry
ret void
}
This shouldn't need the ((zext (%n - 1)) + 1) game, and it should ideally fold
the two memset's together. The issue with %n seems to stem from poor handling
of the original loop.
//===---------------------------------------------------------------------===//