Dan Gohman d04a608a73 Teach SimplifyCFG how to simplify indirectbr instructions.
- Eliminate redundant successors.
 - Convert an indirectbr with one successor into a direct branch.

Also, generalize SimplifyCFG to be able to be run on a function entry block.
It knows quite a few simplifications which are applicable to the entry
block, and it only needs a few checks to avoid trouble with the entry block.

llvm-svn: 111060
2010-08-14 00:29:42 +00:00

52 lines
1.3 KiB
LLVM

; RUN: opt -S -simplifycfg < %s | FileCheck %s
; SimplifyCFG should eliminate redundant indirectbr edges.
; CHECK: indbrtest0
; CHECK: indirectbr i8* %t, [label %BB0, label %BB1, label %BB2]
; CHECK: %x = phi i32 [ 0, %BB0 ], [ 1, %entry ]
declare void @foo()
declare void @A()
declare void @B(i32)
declare void @C()
define void @indbrtest0(i8** %P, i8** %Q) {
entry:
store i8* blockaddress(@indbrtest0, %BB0), i8** %P
store i8* blockaddress(@indbrtest0, %BB1), i8** %P
store i8* blockaddress(@indbrtest0, %BB2), i8** %P
call void @foo()
%t = load i8** %Q
indirectbr i8* %t, [label %BB0, label %BB1, label %BB2, label %BB0, label %BB1, label %BB2]
BB0:
call void @A()
br label %BB1
BB1:
%x = phi i32 [ 0, %BB0 ], [ 1, %entry ], [ 1, %entry ]
call void @B(i32 %x)
ret void
BB2:
call void @C()
ret void
}
; SimplifyCFG should convert the indirectbr into a directbr. It would be even
; better if it removed the branch altogether, but simplifycfdg currently misses
; that because the predecessor is the entry block.
; CHECK: indbrtest1
; CHECK: br label %BB0
define void @indbrtest1(i8** %P, i8** %Q) {
entry:
store i8* blockaddress(@indbrtest1, %BB0), i8** %P
call void @foo()
%t = load i8** %Q
indirectbr i8* %t, [label %BB0, label %BB0]
BB0:
call void @A()
ret void
}