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

例如，我想将两个函数编译成一个 dll：

// test.dll

int plusFive(int value) {
    return value + 5;
}

void printHello() {
    std::cout << "Hello!" << std::endl;
}

并在运行时使用这样的单个函数加载它们：

// main.cc

#include <functional>

int main() {
    std::function<int(int)> func1(loadDllFunc("test.dll", "plusFive"));
    std::function<void()> func2(loadDllFunc("test.dll", "printHello"));
}
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1 回答 1

9

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

  • LoadLibrary- 将指定模块加载到调用进程的地址空间中。返回模块的句柄。
  • 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。

使用此函数加载特定函数并返回一个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loadDllFunc像这样使用：

// 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，可以使用 GCC (4.7.2) 编译 DLL，如下所示：

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

编辑：

我不确定演员loadDllFunc是否给出了正确的类型：

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

它似乎将它投射到__stdcall int (*)(int)应该的时间int (__stdcall *)(int)。

loadDllFunc这是使用辅助解析器类实现的另一种方法。此解决方案将正确地将函数指针转换为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);
}
于 2013-03-06T14:07:55.367 回答