替换 LLVM IR 中的指令
Replacing instructions in LLVM IR
我编写了一段代码,通过创建 Value* 将 LLVM IR 中的添加指令替换为子指令。我正在尝试将 %inc = add i8 %2, 1
替换为 %5 = sub i8 0, %4
。
我的问题是如何使更改出现在 LLVM IR 文件中?我可以在屏幕上打印新值*,但不能在我的 LLVM IR 文件中打印。
for (auto &B : F) {
for (BasicBlock::iterator DI = B.begin(); DI != B.end(); ) {
Instruction *Inst = &*DI++;
if (auto *op = dyn_cast<BinaryOperator>(&*Inst)) {
// Insert at the point where the instruction `op` appears.
IRBuilder<> builder(op);
std::string opcd,opcd_change;
opcd=Inst->getOpcodeName();
// Make a multiply with the same operands as `op`.
srand((unsigned)time(NULL));
if (opcd=="add"){
errs() <<"instruction ";
Inst->print(errs());
//errs() <<'\n'<<"instruction opcode changed"<<opcd_change<<'\n';
errs() <<" instruction opcode "<<opcd<<'\n';
Value *lhs = op->getOperand(0);
Value *rhs = op->getOperand(1);
Instruction* neg = BinaryOperator::CreateNeg(rhs);
errs() <<"instruction opcode changed "<<opcd_change<<'\n';
Instruction* newInst = BinaryOperator::CreateSub(lhs, neg, "test");
errs() << "NewInst:\n" << *newInst << "\n";
ReplaceInstWithInst(op,newInst);
errs()<< "Instruction replaced ";
errs() <<'\n'<<'\n';
}
}
}
}
我无法理解错误的含义。我对 LLVM 还很陌生,所以我不明白它的意思。
**编辑显示使用 ReplaceInstWithInst 后的错误 **
'opt: /home/zainub/llvm/lib/Transforms/Utils/BasicBlockUtils.cpp:211: void llvm::ReplaceInstWithInst(llvm::BasicBlock::InstListType&, llvm::BasicBlock::iterator&, llvm::Instruction*): Assertion `I->getParent() == nullptr && "ReplaceInstWithInst: Instruction already inserted into basic block!"' failed.
#0 0x00000000027abd3a llvm::sys::PrintStackTrace(llvm::raw_ostream&) (/usr/local/bin/opt+0x27abd3a)
#1 0x00000000027ac08e PrintStackTraceSignalHandler(void*) (/usr/local/bin/opt+0x27ac08e)
#2 0x00000000027aa4e6 llvm::sys::RunSignalHandlers() (/usr/local/bin/opt+0x27aa4e6)
#3 0x00000000027ab687 SignalHandler(int) (/usr/local/bin/opt+0x27ab687)
#4 0x00007fbb71d32d10 __restore_rt (/lib/x86_64-linux-gnu/libpthread.so.0+0x10d10)
#5 0x00007fbb71160267 gsignal (/lib/x86_64-linux-gnu/libc.so.6+0x35267)
#6 0x00007fbb71161eca abort (/lib/x86_64-linux-gnu/libc.so.6+0x36eca)
#7 0x00007fbb7115903d (/lib/x86_64-linux-gnu/libc.so.6+0x2e03d)
#8 0x00007fbb711590f2 (/lib/x86_64-linux-gnu/libc.so.6+0x2e0f2)
#9 0x00000000027cfa1c llvm::ReplaceInstWithInst(llvm::SymbolTableList<llvm::Instruction>&, llvm::ilist_iterator<llvm::Instruction>&, llvm::Instruction*) (/usr/local/bin/opt+0x27cfa1c)
#10 0x00000000027cfb2b llvm::ReplaceInstWithInst(llvm::Instruction*, llvm::Instruction*) (/usr/local/bin/opt+0x27cfb2b)
#11 0x00007fbb70f1d027 (anonymous namespace)::Fundep::runOnFunction(llvm::Function&) /home/zainub/llvm/lib/Transforms/Fundep/Fundep.cpp:161:0
#12 0x0000000002246841 llvm::FPPassManager::runOnFunction(llvm::Function&) (/usr/local/bin/opt+0x2246841)
#13 0x00000000022469f8 llvm::FPPassManager::runOnModule(llvm::Module&) (/usr/local/bin/opt+0x22469f8)
#14 0x0000000002246db5 (anonymous namespace)::MPPassManager::runOnModule(llvm::Module&) (/usr/local/bin/opt+0x2246db5)
#15 0x0000000002247539 llvm::legacy::PassManagerImpl::run(llvm::Module&) (/usr/local/bin/opt+0x2247539)
#16 0x0000000002247779 llvm::legacy::PassManager::run(llvm::Module&) (/usr/local/bin/opt+0x2247779)
#17 0x00000000010583e9 main (/usr/local/bin/opt+0x10583e9)
#18 0x00007fbb7114ba40 __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x20a40)
#19 0x0000000001033119 _start (/usr/local/bin/opt+0x1033119)
Stack dump:
0. Program arguments: /usr/local/bin/opt -load /home/zainub/build/lib/LLVMFundep.so -Fundep
1. Running pass 'Function Pass Manager' on module '<stdin>'.
2. Running pass 'Fundep Pass' on function '@main'
已中止(核心已转储)
编辑-2
生成的 IR 有问题
instruction %add115 = add i64 %86, 20 instruction opcode add
NewInst:
%test = sub i64 %86, <badref>
op %add115 = add i64 %86, 20
Instruction replaced %test = sub i64 %86, <badref>
我不确定为什么您的代码段不起作用,但无论如何,将一条指令替换为另一条指令的规范方法是使用 ReplaceInstWithInst.
Instruction
子类 Value
和 BinaryOperation
子类 Instruction
,因此 ReplaceInstWithInst 的第一个参数将只是您的 op
。第二个参数应该是 CreateSub 的 return 值 - 你可以 dyn_cast 它到 Instruction
它很可能会成功。
不过,CreateSub 有可能会 return 一些不是指令的东西,以防编译器设法将你新创建的指令折叠成更简单的东西(例如常量)。如果您担心这种情况,请改用 ReplaceInstWithValue,将其传递给您的 B->getInstList()
和 DI
.
我编写了一段代码,通过创建 Value* 将 LLVM IR 中的添加指令替换为子指令。我正在尝试将 %inc = add i8 %2, 1
替换为 %5 = sub i8 0, %4
。
我的问题是如何使更改出现在 LLVM IR 文件中?我可以在屏幕上打印新值*,但不能在我的 LLVM IR 文件中打印。
for (auto &B : F) {
for (BasicBlock::iterator DI = B.begin(); DI != B.end(); ) {
Instruction *Inst = &*DI++;
if (auto *op = dyn_cast<BinaryOperator>(&*Inst)) {
// Insert at the point where the instruction `op` appears.
IRBuilder<> builder(op);
std::string opcd,opcd_change;
opcd=Inst->getOpcodeName();
// Make a multiply with the same operands as `op`.
srand((unsigned)time(NULL));
if (opcd=="add"){
errs() <<"instruction ";
Inst->print(errs());
//errs() <<'\n'<<"instruction opcode changed"<<opcd_change<<'\n';
errs() <<" instruction opcode "<<opcd<<'\n';
Value *lhs = op->getOperand(0);
Value *rhs = op->getOperand(1);
Instruction* neg = BinaryOperator::CreateNeg(rhs);
errs() <<"instruction opcode changed "<<opcd_change<<'\n';
Instruction* newInst = BinaryOperator::CreateSub(lhs, neg, "test");
errs() << "NewInst:\n" << *newInst << "\n";
ReplaceInstWithInst(op,newInst);
errs()<< "Instruction replaced ";
errs() <<'\n'<<'\n';
}
}
}
}
我无法理解错误的含义。我对 LLVM 还很陌生,所以我不明白它的意思。
**编辑显示使用 ReplaceInstWithInst 后的错误 **
'opt: /home/zainub/llvm/lib/Transforms/Utils/BasicBlockUtils.cpp:211: void llvm::ReplaceInstWithInst(llvm::BasicBlock::InstListType&, llvm::BasicBlock::iterator&, llvm::Instruction*): Assertion `I->getParent() == nullptr && "ReplaceInstWithInst: Instruction already inserted into basic block!"' failed.
#0 0x00000000027abd3a llvm::sys::PrintStackTrace(llvm::raw_ostream&) (/usr/local/bin/opt+0x27abd3a)
#1 0x00000000027ac08e PrintStackTraceSignalHandler(void*) (/usr/local/bin/opt+0x27ac08e)
#2 0x00000000027aa4e6 llvm::sys::RunSignalHandlers() (/usr/local/bin/opt+0x27aa4e6)
#3 0x00000000027ab687 SignalHandler(int) (/usr/local/bin/opt+0x27ab687)
#4 0x00007fbb71d32d10 __restore_rt (/lib/x86_64-linux-gnu/libpthread.so.0+0x10d10)
#5 0x00007fbb71160267 gsignal (/lib/x86_64-linux-gnu/libc.so.6+0x35267)
#6 0x00007fbb71161eca abort (/lib/x86_64-linux-gnu/libc.so.6+0x36eca)
#7 0x00007fbb7115903d (/lib/x86_64-linux-gnu/libc.so.6+0x2e03d)
#8 0x00007fbb711590f2 (/lib/x86_64-linux-gnu/libc.so.6+0x2e0f2)
#9 0x00000000027cfa1c llvm::ReplaceInstWithInst(llvm::SymbolTableList<llvm::Instruction>&, llvm::ilist_iterator<llvm::Instruction>&, llvm::Instruction*) (/usr/local/bin/opt+0x27cfa1c)
#10 0x00000000027cfb2b llvm::ReplaceInstWithInst(llvm::Instruction*, llvm::Instruction*) (/usr/local/bin/opt+0x27cfb2b)
#11 0x00007fbb70f1d027 (anonymous namespace)::Fundep::runOnFunction(llvm::Function&) /home/zainub/llvm/lib/Transforms/Fundep/Fundep.cpp:161:0
#12 0x0000000002246841 llvm::FPPassManager::runOnFunction(llvm::Function&) (/usr/local/bin/opt+0x2246841)
#13 0x00000000022469f8 llvm::FPPassManager::runOnModule(llvm::Module&) (/usr/local/bin/opt+0x22469f8)
#14 0x0000000002246db5 (anonymous namespace)::MPPassManager::runOnModule(llvm::Module&) (/usr/local/bin/opt+0x2246db5)
#15 0x0000000002247539 llvm::legacy::PassManagerImpl::run(llvm::Module&) (/usr/local/bin/opt+0x2247539)
#16 0x0000000002247779 llvm::legacy::PassManager::run(llvm::Module&) (/usr/local/bin/opt+0x2247779)
#17 0x00000000010583e9 main (/usr/local/bin/opt+0x10583e9)
#18 0x00007fbb7114ba40 __libc_start_main (/lib/x86_64-linux-gnu/libc.so.6+0x20a40)
#19 0x0000000001033119 _start (/usr/local/bin/opt+0x1033119)
Stack dump:
0. Program arguments: /usr/local/bin/opt -load /home/zainub/build/lib/LLVMFundep.so -Fundep
1. Running pass 'Function Pass Manager' on module '<stdin>'.
2. Running pass 'Fundep Pass' on function '@main'
已中止(核心已转储)
编辑-2
生成的 IR 有问题
instruction %add115 = add i64 %86, 20 instruction opcode add
NewInst:
%test = sub i64 %86, <badref>
op %add115 = add i64 %86, 20
Instruction replaced %test = sub i64 %86, <badref>
我不确定为什么您的代码段不起作用,但无论如何,将一条指令替换为另一条指令的规范方法是使用 ReplaceInstWithInst.
Instruction
子类 Value
和 BinaryOperation
子类 Instruction
,因此 ReplaceInstWithInst 的第一个参数将只是您的 op
。第二个参数应该是 CreateSub 的 return 值 - 你可以 dyn_cast 它到 Instruction
它很可能会成功。
不过,CreateSub 有可能会 return 一些不是指令的东西,以防编译器设法将你新创建的指令折叠成更简单的东西(例如常量)。如果您担心这种情况,请改用 ReplaceInstWithValue,将其传递给您的 B->getInstList()
和 DI
.