从DLL动态加载任意函数到std::function

Question

如何使用单个函数将任意动态链接库 (dll)函数加载到std::function对象中?

例如，我想将两个函数编译成dll:

// test.dll
int plusFive(int value) {
    return value + 5;
}
void printHello() {
    std::cout << "Hello!" << std::endl;
}

并在运行时使用一个函数加载它们，像这样:

// main.cc
#include <functional>
int main() {
    std::function<int(int)> func1(loadDllFunc("test.dll", "plusFive"));
    std::function<void()> func2(loadDllFunc("test.dll", "printHello"));
}

Answer 1

使用windows.h中提供的WinAPI函数(描述取自MSDN Dev Center)

• LoadLibrary -将指定模块加载到调用进程的地址空间中。返回一个句柄到模块。
• GetProcAddress -从指定的动态链接库(DLL)中检索导出函数或变量的地址。返回导出函数或变量的地址。

使用这个函数来加载一个特定的函数并返回一个std::function对象:

// main.cc
#include <iostream>
#include <string>
#include <functional>
#include <windows.h>
template <typename T>
std::function<T> loadDllFunc(const std::string& dllName, const std::string& funcName) {
    // Load DLL.
    HINSTANCE hGetProcIDDLL = LoadLibrary(dllName.c_str());
    // Check if DLL is loaded.
    if (hGetProcIDDLL == NULL) {
        std::cerr << "Could not load DLL "" << dllName << """ << std::endl;
        exit(EXIT_FAILURE);
    }
    // Locate function in DLL.
    FARPROC lpfnGetProcessID = GetProcAddress(hGetProcIDDLL, funcName.c_str());
    // Check if function was located.
    if (!lpfnGetProcessID) {
        std::cerr << "Could not locate the function "" << funcName << "" in DLL"" << dllName << """ << std::endl;
        exit(EXIT_FAILURE);
    }
    // Create function object from function pointer.
    std::function<T> func(reinterpret_cast<__stdcall T*>(lpfnGetProcessID));
    return func;
}

DLL源代码应该这样写:

// test.cc (test.dll)
#include <iostream>
// Declare function prototypes with "extern C" to prevent name mangling.
// Declare functions using __declspec(dllexport) to signify the intent to export.
extern "C" {
    __declspec(dllexport) int __stdcall plusFive(int);
    __declspec(dllexport) void __stdcall printHello();
}
int plusFive(int value) {
    return value + 5;
}
void printHello() {
    std::cout << "Hello!" << std::endl;
}

然后像这样使用loadDllFunc:

// main.cc
int main() {
    auto func1 = loadDllFunc<int(int)>("test.dll", "plusFive");
    auto func2 = loadDllFunc<void()>("test.dll", "printHello");
    std::cout << "Result of func1: " << func1(1) << std::endl;
    func2();
}

输出:

Result of func1: 6
Hello!

作为旁注，DLL可以像这样使用GCC(4.7.2)编译:

g++ -shared -o test.dll test.cc -std=c++11

编辑:

我不确定loadDllFunc中的转换是否给出了正确的类型:

std::function<T> func(reinterpret_cast<__stdcall T*>(lpfnGetProcessID));

当它应该是int (__stdcall *)(int)时，它似乎将其强制转换为__stdcall int (*)(int)。

下面是使用辅助解析器类实现loadDllFunc的另一种方法。此解决方案将正确地将函数指针强制转换为int (__stdcall *)(int)。

template <typename T>
struct TypeParser {};
template <typename Ret, typename... Args>
struct TypeParser<Ret(Args...)> {
    static std::function<Ret(Args...)> createFunction(const FARPROC lpfnGetProcessID) {
        return std::function<Ret(Args...)>(reinterpret_cast<Ret (__stdcall *)(Args...)>(lpfnGetProcessID));
    }
};
template <typename T>
std::function<T> loadDllFunc(const std::string& dllName, const std::string& funcName) {
    // Load DLL.
    HINSTANCE hGetProcIDDLL = LoadLibrary(dllName.c_str());
    // Check if DLL is loaded.
    if (hGetProcIDDLL == NULL) {
        std::cerr << "Could not load DLL "" << dllName << """ << std::endl;
        exit(EXIT_FAILURE);
    }
    // Locate function in DLL.
    FARPROC lpfnGetProcessID = GetProcAddress(hGetProcIDDLL, funcName.c_str());
    // Check if function was located.
    if (!lpfnGetProcessID) {
        std::cerr << "Could not locate the function "" << funcName << "" in DLL"" << dllName << """ << std::endl;
        exit(EXIT_FAILURE);
    }
    // Create function object from function pointer.
    return TypeParser<T>::createFunction(lpfnGetProcessID);
}