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到一个dll中:
// test.dll
int plusFive(int value) {
return value + 5;
}
void printHello() {
std::cout << "Hello!" << std::endl;
}
并在运行时使用单个函数加载它们,如下所示:
// main.cc
#include <functional>
int main() {
std::function<int(int)> func1(loadDllFunc("test.dll", "plusFive"));
std::function<void()> func2(loadDllFunc("test.dll", "printHello"));
}
最佳答案
使用 windows.h
中提供的 WinAPI 函数(描述取自 MSDN Dev Center)。
LoadLibrary
- 将指定的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) 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 可以使用 GCC (4.7.2) 进行编译,如下所示:
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));
它似乎将其转换为 __stdcall int (*)(int)
,而它应该是 int (__stdcall *)(int)
。
这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);
}
关于C++ 将任意函数从 DLL 动态加载到 std::function,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15249465/