我想在LLVM中创建一个函数,它是一个只有函数调用foo(idx,mn)的适配器。foo的功能原型是void foo(unsigned char, const char*)
。
// adapter Function with only a function call foo(idx, mn)
llvm::Function* createCallFun(llvm::Module* M, llvm::Function* exit_f) {
llvm::LLVMContext& Ctx = M->getContext();
llvm::Function* foo_f = foo_prototype(Ctx, M);
llvm::Constant* c = M->getOrInsertFunction("__call_fun", FunctionType::getVoidTy(Ctx), llvm::Type::getInt32Ty(Ctx), llvm::Type::getInt32Ty(Ctx), NULL);
llvm::Function* call_fun_f = llvm::cast<llvm::Function>(c);
llvm::BasicBlock* entry = llvm::BasicBlock::Create(llvm::getGlobalContext(), "entry", call_fun_f);
llvm::IRBuilder<> builder(entry);
llvm::Function::arg_iterator args = call_fun_f->arg_begin();
llvm::Value* idx = &*args++;
idx->setName("idx");
llvm::Value* mn = &*args++;
mn->setName("mn");
llvm::Value* greater = builder.CreateICmpSGE(idx, mn, "tmp");
std::vector<llvm::Value*> fun_args;
fun_args.push_back(greater);
fun_args.push_back(err_msg);
builder.CreateCall(foo_f, fun_args);
return call_fun_f;
}
然后我得到了一个错误:
lib/IR/Instructions.cpp:245: void::CallInst::init(llvm::FunctionType*,llvm::Value*,llvm::ArrayRef,llvm::ArrayRef >,const llvm::Twine&):断言`(i >= FTy->getNumParams() \ FTy->getParamType(i) == Argsi->getType()) &“调用带有错误签名的函数!”失败。
foo的第一个参数似乎存在类型不匹配。如何将值greater
转换为unsigned char
类型?
发布于 2016-12-18 12:55:29
我通过使用greater
和CreateZExt
强制转换来修正这个错误。
llvm::Value *castuchar =
builder.CreateZExt(greater, llvm::Type::getInt8Ty(Ctx), "tmp1");
https://stackoverflow.com/questions/41207523
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