[SimplifyCFG] NFC; remove bogus test case

Even if one bit is defined, the code is not clear what it is suppose to do.

The test wants to assert that some bits are undef, but that's not what the IR does and I don't think it's even possible to do that in any meaningful way. It was added in D12497, so @reames might want to double check.

Differential Revision: https://reviews.llvm.org/D60859

llvm-svn: 362499

1 files changed, 0 insertions, 26 deletions

diff --git a/llvm/test/Transforms/SimplifyCFG/switch-dead-default.ll b/llvm/test/Transforms/SimplifyCFG/switch-dead-default.ll
index e5c2ef65b31..e14eab98c69 100644
--- a/llvm/test/Transforms/SimplifyCFG/switch-dead-default.ll
+++ b/llvm/test/Transforms/SimplifyCFG/switch-dead-default.ll
@@ -151,29 +151,3 @@ default:
   call void @foo(i32 2)
   ret void
 }
-
-;; All but one bit known undef
-;; Note: This is currently testing an optimization which doesn't trigger. The
-;; case this is protecting against is that a bit could be assumed both zero 
-;; *or* one given we know it's undef.  ValueTracking doesn't do this today,
-;; but it doesn't hurt to confirm.
-define void @test8(i8 %a) {
-; CHECK-LABEL: @test8(
-; CHECK: switch i8
-  %and = and i8 %a, 254
-  %cmp = icmp eq i8 %and, undef
-  call void @llvm.assume(i1 %cmp)
-  switch i8 %a, label %default [i8 255, label %true
-                                i8 254, label %false]
-true:
-  call void @foo(i32 1)
-  ret void
-false:
-  call void @foo(i32 3)
-  ret void
-default:
-  call void @foo(i32 2)
-  ret void
-}
-
-declare void @llvm.assume(i1)