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merge two tests and convert to filecheck.

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git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@84671 91177308-0d34-0410-b5e6-96231b3b80d8
This commit is contained in:
Chris Lattner 2009-10-20 20:33:46 +00:00
parent 3229b0bcf1
commit 7f2076023a
2 changed files with 55 additions and 45 deletions
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17
test/Transforms/SCCP/basictest.ll
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@ -1,17 +0,0 @@
; This is a basic sanity check for constant propogation. The add instruction
; should be eliminated.
; RUN: opt < %s -sccp -S | not grep add
define i32 @test(i1 %B) {
br i1 %B, label %BB1, label %BB2
BB1: ; preds = %0
%Val = add i32 0, 0 ; <i32> [#uses=1]
br label %BB3
BB2: ; preds = %0
br label %BB3
BB3: ; preds = %BB2, %BB1
%Ret = phi i32 [ %Val, %BB1 ], [ 1, %BB2 ] ; <i32> [#uses=1]
ret i32 %Ret
}

83
test/Transforms/SCCP/sccptest.ll
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@ -1,31 +1,58 @@
; This is the test case taken from appel's book that illustrates a hard case
; that SCCP gets right. BB3 should be completely eliminated.
;
; RUN: opt < %s -sccp -constprop -dce -simplifycfg | \
; RUN: llvm-dis | not grep BB3
; RUN: opt < %s -sccp -S | FileCheck %s
define i32 @testfunction(i32 %i0, i32 %j0) {
BB1:
br label %BB2
BB2: ; preds = %BB7, %BB1
%j2 = phi i32 [ %j4, %BB7 ], [ 1, %BB1 ] ; <i32> [#uses=2]
%k2 = phi i32 [ %k4, %BB7 ], [ 0, %BB1 ] ; <i32> [#uses=4]
%kcond = icmp slt i32 %k2, 100 ; <i1> [#uses=1]
br i1 %kcond, label %BB3, label %BB4
BB3: ; preds = %BB2
%jcond = icmp slt i32 %j2, 20 ; <i1> [#uses=1]
br i1 %jcond, label %BB5, label %BB6
BB4: ; preds = %BB2
ret i32 %j2
BB5: ; preds = %BB3
%k3 = add i32 %k2, 1 ; <i32> [#uses=1]
br label %BB7
BB6: ; preds = %BB3
%k5 = add i32 %k2, 1 ; <i32> [#uses=1]
br label %BB7
BB7: ; preds = %BB6, %BB5
%j4 = phi i32 [ 1, %BB5 ], [ %k2, %BB6 ] ; <i32> [#uses=1]
%k4 = phi i32 [ %k3, %BB5 ], [ %k5, %BB6 ] ; <i32> [#uses=1]
br label %BB2
; This is a basic sanity check for constant propagation. The add instruction
; should be eliminated.
define i32 @test1(i1 %B) {
br i1 %B, label %BB1, label %BB2
BB1: ; preds = %0
%Val = add i32 0, 0 ; <i32> [#uses=1]
br label %BB3
BB2: ; preds = %0
br label %BB3
BB3: ; preds = %BB2, %BB1
%Ret = phi i32 [ %Val, %BB1 ], [ 1, %BB2 ] ; <i32> [#uses=1]
ret i32 %Ret
; CHECK: @test1
; CHECK: %Ret = phi i32 [ 0, %BB1 ], [ 1, %BB2 ]
}
; This is the test case taken from appel's book that illustrates a hard case
; that SCCP gets right.
;
define i32 @test2(i32 %i0, i32 %j0) {
; CHECK: @test2
BB1:
br label %BB2
BB2:
%j2 = phi i32 [ %j4, %BB7 ], [ 1, %BB1 ]
%k2 = phi i32 [ %k4, %BB7 ], [ 0, %BB1 ]
%kcond = icmp slt i32 %k2, 100
br i1 %kcond, label %BB3, label %BB4
BB3:
%jcond = icmp slt i32 %j2, 20
br i1 %jcond, label %BB5, label %BB6
; CHECK: BB3:
; CHECK-NEXT: br i1 true, label %BB5, label %BB6
BB4:
ret i32 %j2
; CHECK: BB4:
; CHECK-NEXT: ret i32 1
BB5:
%k3 = add i32 %k2, 1
br label %BB7
BB6:
%k5 = add i32 %k2, 1
br label %BB7
; CHECK: BB6:
; CHECK-NEXT: br label %BB7
BB7:
%j4 = phi i32 [ 1, %BB5 ], [ %k2, %BB6 ]
%k4 = phi i32 [ %k3, %BB5 ], [ %k5, %BB6 ]
br label %BB2
; CHECK: BB7:
; CHECK-NEXT: %k4 = phi i32 [ %k3, %BB5 ], [ undef, %BB6 ]
; CHECK-NEXT: br label %BB2
}