首页 > 解决方案 > llvm 以 char * 作为参数调用外部函数

问题描述

我想在创建的调用中使用 char * 作为参数调用外部函数,但我一直在苦苦挣扎。

我尝试查看文档,但它只显示了如何传递常量整数。

这是代码的相关部分:

// not sure if the line below is correct. I am trying to call an external function: FPRCLAP_path(char * string)
std::vector<llvm::Type*> arg_types = {llvm::Type::getInt8PtrTy(context)};

llvm::FunctionType *function_type = llvm::FunctionType::get(llvm::Type::getVoidTy(context), arg_types, false);
llvm::FunctionCallee instrumentation_function = function.getParent()->getOrInsertFunction("FPRCLAP_path", function_type);

llvm::IRBuilder<> builder(&instruction);
builder.SetInsertPoint(&basic_block, ++builder.GetInsertPoint());
// I want to pass a string for instrumentation but I am not sure how to do it using llvm. I think I am supposed to allocate a string or reference an existing string.
llvm::ArrayRef<llvm::Value*> arguments = {???};
builder.CreateCall(instrumentation_function, arguments);

标签: c++llvmllvm-clangllvm-c++-api

解决方案


考虑你的代码

llvm::IRBuilder<> builder(&instruction);
builder.SetInsertPoint(&basic_block, ++builder.GetInsertPoint());
// I want to pass a string for instrumentation but I am not sure how to do it using llvm. I think I am supposed to allocate a string or reference an existing string.
llvm::ArrayRef<llvm::Value*> arguments = {???};
builder.CreateCall(instrumentation_function, arguments);

下面的代码片段应该做你想做的

如果运行,LLVM 优化将使字符串本地化。如果您在没有优化的情况下生成 IR,您可能会有点草率。

//str is a char*
 llvm::Value *strPointer = program->builder.CreateGlobalStringPtr(str);
//Using a Vector instead of ArrayRef
 const std::vector<llvm::Value *> args{strPointer};
 builder.CreateCall(instrumentation_function, args);

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