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More testcases for SCCP
git-svn-id: https://llvm.org/svn/llvm-project/llvm/trunk@2444 91177308-0d34-0410-b5e6-96231b3b80d8
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28
test/Transforms/SCCP/2002-05-02-EdgeFailure.ll
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28
test/Transforms/SCCP/2002-05-02-EdgeFailure.ll
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; edgefailure - This function illustrates how SCCP is not doing it's job. This
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; function should be optimized almost completely away: the loop should be
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; analyzed to detect that the body executes exactly once, and thus the branch
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; can be eliminated and code becomes trivially dead. This is distilled from a
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; real benchmark (mst from Olden benchmark, MakeGraph function). When SCCP is
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; fixed, this should be eliminated by a single SCCP application.
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;
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; RUN: if as < %s | opt -sccp | dis | grep loop
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; RUN: then exit 1
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; RUN: else exit 0
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; RUN: fi
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int *"test"()
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begin
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bb1:
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%A = malloc int
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br label %bb2
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bb2:
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%i = phi int [ %i2, %bb2 ], [ 0, %bb1 ] ;; Always 0
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%i2 = add int %i, 1 ;; Always 1
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store int %i, int *%A
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%loop = setle int %i2, 0 ;; Always false
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br bool %loop, label %bb2, label %bb3
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bb3:
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ret int * %A
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end
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20
test/Transforms/SCCP/basictest.ll
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test/Transforms/SCCP/basictest.ll
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; This is a basic sanity check for constant propogation. The add instruction
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; should be eliminated.
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; RUN: if as < %s | opt -sccp | dis | grep add
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; RUN: then exit 1
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; RUN: else exit 0
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; RUN: fi
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int "test"(bool %B)
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begin
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br bool %B, label %BB1, label %BB2
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BB1:
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%Val = add int 0, 0
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br label %BB3
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BB2:
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br label %BB3
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BB3:
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%Ret = phi int [%Val, %BB1], [1, %BB2]
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ret int %Ret
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end
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39
test/Transforms/SCCP/sccptest.ll
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39
test/Transforms/SCCP/sccptest.ll
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; This is the test case taken from appel's book that illustrates a hard case
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; that SCCP gets right. BB3 should be completely eliminated.
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;
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; RUN: if as < %s | opt -sccp -dce | dis | grep BB3
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; RUN: then exit 1
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; RUN: else exit 0
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; RUN: fi
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int "test function"(int %i0, int %j0)
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begin
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BB1:
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br label %BB2
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BB2:
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%j2 = phi int [%j4, %BB7], [1, %BB1]
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%k2 = phi int [%k4, %BB7], [0, %BB1]
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%kcond = setlt int %k2, 100
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br bool %kcond, label %BB3, label %BB4
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BB3:
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%jcond = setlt int %j2, 20
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br bool %jcond, label %BB5, label %BB6
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BB4:
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ret int %j2
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BB5:
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%k3 = add int %k2, 1
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br label %BB7
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BB6:
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%k5 = add int %k2, 1
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br label %BB7
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BB7:
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%j4 = phi int [1, %BB5], [%k2, %BB6]
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%k4 = phi int [%k3, %BB5], [%k5, %BB6]
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br label %BB2
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end
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