从 LLVM 添加到 C++ lambda 的映射
add mapping to C++ lambda from LLVM
我有一组包装 C++ 方法调用的 lambda,我想用 LLVM 调用它们。我的尝试似乎遗漏了一些东西,即使在声明了类型并将全局映射添加到 lambda 之后,我仍然收到 LLVM 错误。重现我尝试的最小代码是:
#include "llvm/ExecutionEngine/GenericValue.h"
#include "llvm/ExecutionEngine/Interpreter.h"
#include "llvm/IR/Constants.h"
#include "llvm/IR/IRBuilder.h"
#include "llvm/Support/ManagedStatic.h"
#include "llvm/Support/TargetSelect.h"
using namespace llvm;
int main() {
InitializeNativeTarget();
LLVMContext Context;
std::unique_ptr<Module> Owner = make_unique<Module>("SomeModule", Context);
Module *M = Owner.get();
FunctionType *lambdaFT = FunctionType::get(Type::getInt32Ty(Context), false);
Function *lambdaFN = Function::Create(lambdaFT, Function::ExternalLinkage, "lambda", Owner.get());
//this is what the original question had
//auto lambdaBody = []() { return 100; };
//this is an edit after Johannes Schaub's answer
int32_t ( *lambdaBody)() = +[]() { return 100; };
Function *mainF = cast<Function>(M->getOrInsertFunction("main", Type::getInt32Ty(Context), (Type *) 0));
BasicBlock *BB = BasicBlock::Create(Context, "EntryBlock", mainF);
IRBuilder<> builder(BB);
CallInst *lambdaRes = builder.CreateCall(lambdaFN, std::vector<Value *>(), "lambdaRetVar");
builder.CreateRet(lambdaRes);
ExecutionEngine *EE = EngineBuilder(std::move(Owner)).create();
EE->addGlobalMapping(lambdaFN, &lambdaBody);
outs() << "We just constructed this LLVM module:\n\n" << *M;
outs() << "\n\nRunning main: ";
std::vector<GenericValue> noargs;
GenericValue gv = EE->runFunction(mainF, noargs);
outs() << "Result: " << gv.IntVal << "\n";
llvm_shutdown();
delete EE;
return 0;
}
这导致输出:
We just constructed this LLVM module:
; ModuleID = 'SomeModule'
declare i32 @lambda()
define i32 @main() {
EntryBlock:
%lambdaRetVar = call i32 @lambda()
ret i32 %lambdaRetVar
}
Running main:
LLVM ERROR: Tried to execute an unknown external function: lambda
我做错了什么?
使用 LLVM 3.7.0
(标签不存在)
你的 lambda 体是 class。您必须传递其函数调用运算符的地址,您可以将其转换为函数指针:auto lambdaBody = +[]() { return 100; };
并将其作为 void*
:EE->addGlobalMapping(lambdaFN, reinterpret_cast<void*>(lambdaBody));
传递。
Johannes Schaub
关于 lambda 的回答很有帮助,但事实证明 addGlobalMapping
并没有按照我的想法行事。还缺少一些东西。
最重要的是:
EE->InstallLazyFunctionCreator([&](const std::string &fnName) -> void * {
if (fnName == "lambda") { return reinterpret_cast<void *>(lambdaBody); }
return nullptr;
});
InstallLazyFunctionCreator
用于函数无法解析的时候。它以函数名称调用,作为响应,如果未找到此函数 returns nullptr
或要执行的函数。
这是完整的代码清单,其中添加了一些内容(并非全部都是必需的):
#include "llvm/ExecutionEngine/GenericValue.h"
#include "llvm/ExecutionEngine/Interpreter.h"
#include "llvm/ExecutionEngine/RTDyldMemoryManager.h"
//must include to avoid "JIT has not been linked in" when creating the ExecutionEngine - https://llvm.org/bugs/show_bug.cgi?id=22910
#include "llvm/ExecutionEngine/MCJIT.h"
#include "llvm/IR/Constants.h"
#include "llvm/IR/IRBuilder.h"
#include "llvm/Support/ManagedStatic.h"
#include "llvm/Support/TargetSelect.h"
#include <iostream>
using namespace llvm;
int proxy() { return 1000; }
int main() {
InitializeNativeTarget();
llvm::InitializeNativeTargetAsmPrinter();
llvm::InitializeNativeTargetAsmParser();
LLVMContext Context;
std::unique_ptr<Module> Owner = make_unique<Module>("SomeModule", Context);
Module *M = Owner.get();
FunctionType *lambdaFT = FunctionType::get(Type::getInt32Ty(Context), false);
Function *lambdaFN = Function::Create(lambdaFT, Function::ExternalLinkage, "lambda", Owner.get());
int32_t ( *lambdaBody)() = +[]() { return 9099899; };
assert(9099899 == lambdaBody()); //make sure we can call it
Function *mainF = cast<Function>(M->getOrInsertFunction("main", Type::getInt32Ty(Context), (Type *) 0));
BasicBlock *BB = BasicBlock::Create(Context, "EntryBlock", mainF);
IRBuilder<> builder(BB);
CallInst *lambdaRes = builder.CreateCall(lambdaFN, std::vector<Value *>(), "lambdaRetVar");
builder.CreateRet(lambdaRes);
EngineBuilder eb(std::move(Owner));
std::unique_ptr<RTDyldMemoryManager> MM;
std::string Error;
ExecutionEngine *EE = eb.setEngineKind(EngineKind::JIT)
.setMCJITMemoryManager(std::move(MM))
.setErrorStr(&Error)
.setOptLevel(CodeGenOpt::None)
.setCodeModel(CodeModel::JITDefault)
.setRelocationModel(Reloc::Default)
//.setMArch(MArch)
.setMCPU(sys::getHostCPUName())
//.setMAttrs(MAttrs)
.create();
std::cout << Error << std::endl;
EE->InstallLazyFunctionCreator([&](const std::string &fnName) -> void * {
if (fnName == "lambda") { return reinterpret_cast<void *>(lambdaBody); }
return nullptr;
});
EE->finalizeObject();
outs() << "We just constructed this LLVM module:\n\n" << *M;
outs() << "\n\nRunning main: ";
std::vector<GenericValue> noargs;
GenericValue gv = EE->runFunction(mainF, noargs);
outs() << "Result: " << gv.IntVal << "\n";
assert(gv.IntVal == 9099899);
llvm_shutdown();
delete EE;
return 0;
}
我有一组包装 C++ 方法调用的 lambda,我想用 LLVM 调用它们。我的尝试似乎遗漏了一些东西,即使在声明了类型并将全局映射添加到 lambda 之后,我仍然收到 LLVM 错误。重现我尝试的最小代码是:
#include "llvm/ExecutionEngine/GenericValue.h"
#include "llvm/ExecutionEngine/Interpreter.h"
#include "llvm/IR/Constants.h"
#include "llvm/IR/IRBuilder.h"
#include "llvm/Support/ManagedStatic.h"
#include "llvm/Support/TargetSelect.h"
using namespace llvm;
int main() {
InitializeNativeTarget();
LLVMContext Context;
std::unique_ptr<Module> Owner = make_unique<Module>("SomeModule", Context);
Module *M = Owner.get();
FunctionType *lambdaFT = FunctionType::get(Type::getInt32Ty(Context), false);
Function *lambdaFN = Function::Create(lambdaFT, Function::ExternalLinkage, "lambda", Owner.get());
//this is what the original question had
//auto lambdaBody = []() { return 100; };
//this is an edit after Johannes Schaub's answer
int32_t ( *lambdaBody)() = +[]() { return 100; };
Function *mainF = cast<Function>(M->getOrInsertFunction("main", Type::getInt32Ty(Context), (Type *) 0));
BasicBlock *BB = BasicBlock::Create(Context, "EntryBlock", mainF);
IRBuilder<> builder(BB);
CallInst *lambdaRes = builder.CreateCall(lambdaFN, std::vector<Value *>(), "lambdaRetVar");
builder.CreateRet(lambdaRes);
ExecutionEngine *EE = EngineBuilder(std::move(Owner)).create();
EE->addGlobalMapping(lambdaFN, &lambdaBody);
outs() << "We just constructed this LLVM module:\n\n" << *M;
outs() << "\n\nRunning main: ";
std::vector<GenericValue> noargs;
GenericValue gv = EE->runFunction(mainF, noargs);
outs() << "Result: " << gv.IntVal << "\n";
llvm_shutdown();
delete EE;
return 0;
}
这导致输出:
We just constructed this LLVM module:
; ModuleID = 'SomeModule'
declare i32 @lambda()
define i32 @main() {
EntryBlock:
%lambdaRetVar = call i32 @lambda()
ret i32 %lambdaRetVar
}
Running main:
LLVM ERROR: Tried to execute an unknown external function: lambda
我做错了什么?
使用 LLVM 3.7.0
(标签不存在)
你的 lambda 体是 class。您必须传递其函数调用运算符的地址,您可以将其转换为函数指针:auto lambdaBody = +[]() { return 100; };
并将其作为 void*
:EE->addGlobalMapping(lambdaFN, reinterpret_cast<void*>(lambdaBody));
传递。
Johannes Schaub
关于 lambda 的回答很有帮助,但事实证明 addGlobalMapping
并没有按照我的想法行事。还缺少一些东西。
最重要的是:
EE->InstallLazyFunctionCreator([&](const std::string &fnName) -> void * {
if (fnName == "lambda") { return reinterpret_cast<void *>(lambdaBody); }
return nullptr;
});
InstallLazyFunctionCreator
用于函数无法解析的时候。它以函数名称调用,作为响应,如果未找到此函数 returns nullptr
或要执行的函数。
这是完整的代码清单,其中添加了一些内容(并非全部都是必需的):
#include "llvm/ExecutionEngine/GenericValue.h"
#include "llvm/ExecutionEngine/Interpreter.h"
#include "llvm/ExecutionEngine/RTDyldMemoryManager.h"
//must include to avoid "JIT has not been linked in" when creating the ExecutionEngine - https://llvm.org/bugs/show_bug.cgi?id=22910
#include "llvm/ExecutionEngine/MCJIT.h"
#include "llvm/IR/Constants.h"
#include "llvm/IR/IRBuilder.h"
#include "llvm/Support/ManagedStatic.h"
#include "llvm/Support/TargetSelect.h"
#include <iostream>
using namespace llvm;
int proxy() { return 1000; }
int main() {
InitializeNativeTarget();
llvm::InitializeNativeTargetAsmPrinter();
llvm::InitializeNativeTargetAsmParser();
LLVMContext Context;
std::unique_ptr<Module> Owner = make_unique<Module>("SomeModule", Context);
Module *M = Owner.get();
FunctionType *lambdaFT = FunctionType::get(Type::getInt32Ty(Context), false);
Function *lambdaFN = Function::Create(lambdaFT, Function::ExternalLinkage, "lambda", Owner.get());
int32_t ( *lambdaBody)() = +[]() { return 9099899; };
assert(9099899 == lambdaBody()); //make sure we can call it
Function *mainF = cast<Function>(M->getOrInsertFunction("main", Type::getInt32Ty(Context), (Type *) 0));
BasicBlock *BB = BasicBlock::Create(Context, "EntryBlock", mainF);
IRBuilder<> builder(BB);
CallInst *lambdaRes = builder.CreateCall(lambdaFN, std::vector<Value *>(), "lambdaRetVar");
builder.CreateRet(lambdaRes);
EngineBuilder eb(std::move(Owner));
std::unique_ptr<RTDyldMemoryManager> MM;
std::string Error;
ExecutionEngine *EE = eb.setEngineKind(EngineKind::JIT)
.setMCJITMemoryManager(std::move(MM))
.setErrorStr(&Error)
.setOptLevel(CodeGenOpt::None)
.setCodeModel(CodeModel::JITDefault)
.setRelocationModel(Reloc::Default)
//.setMArch(MArch)
.setMCPU(sys::getHostCPUName())
//.setMAttrs(MAttrs)
.create();
std::cout << Error << std::endl;
EE->InstallLazyFunctionCreator([&](const std::string &fnName) -> void * {
if (fnName == "lambda") { return reinterpret_cast<void *>(lambdaBody); }
return nullptr;
});
EE->finalizeObject();
outs() << "We just constructed this LLVM module:\n\n" << *M;
outs() << "\n\nRunning main: ";
std::vector<GenericValue> noargs;
GenericValue gv = EE->runFunction(mainF, noargs);
outs() << "Result: " << gv.IntVal << "\n";
assert(gv.IntVal == 9099899);
llvm_shutdown();
delete EE;
return 0;
}