C ++从DLL动态加载任意函数到std :: function [英] C++ Dynamically load arbitrary function from DLL into std::function
问题描述
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? / p>
例如,我想将两个函数编译为dll:
// test.dll
int plusFive(int value){
返回值+ 5;
}
void printHello(){
std :: cout<< 你好! << std :: endl;
}
并使用以下单个函数在运行时加载它们:
// main.cc
#include< functional>
int main(){
std :: function< int(int)> func1(loadDllFunc( test.dll, plusFive)));
std :: function< void()> func2(loadDllFunc( test.dll, printHello));
}
使用 WinAPI 功能在 windows.h
中提供(描述取自 MSDN开发中心)。
-
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中检索导出的函数或变量的地址( DLL)。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 std :: function
对象:
// main.cc
#include< iostream>
#include< string>
#include< functional>
#include< windows.h>
模板< typename T>
std :: function< T> loadDllFunc(const std :: string& dllName,const std :: string& funcName){
//加载DLL。
HINSTANCE hGetProcIDDLL = LoadLibrary(dllName.c_str());
//检查DLL是否已加载。
if(hGetProcIDDLL == NULL){
std :: cerr<< 无法加载DLL \<< dllName<< \<< std :: endl;
出口(EXIT_FAILURE);
}
//在DLL中定位函数。
FARPROC lpfnGetProcessID = GetProcAddress(hGetProcIDDLL,funcName.c_str());
//检查是否找到函数。
if(!lpfnGetProcessID){
std :: cerr<< 无法在DLL中找到函数\ funcName << \ \ <<< dllName << \ < std :: endl;
出口(EXIT_FAILURE);
}
//从函数指针创建函数对象。
std :: function< T> func(reinterpret_cast< __ stdcall T *>(lpfnGetProcessID));
返回函数;
}
DLL源应该这样写:
// test.cc(test.dll)
#include< iostream>
//使用 extern C声明函数原型,以防止名称篡改。
//使用__declspec(dllexport)声明函数以表示导出意图。
extern C {
__declspec(dllexport)int __stdcall plusFive(int);
__declspec(dllexport)void __stdcall printHello();
}
int plusFive(int value){
返回值+ 5;
}
void printHello(){
std :: cout<< 你好! << std :: endl;
}
然后使用 loadDllFunc
像这样:
// main.cc
int main(){
自动功能1 = loadDllFunc< int(int)>( test.dll, plusFive);
auto func2 = loadDllFunc< void()>( test.dll, printHello);
std :: cout<< func1的结果:<< func1(1)<< std :: endl;
func2();
}
输出:
func1的结果:6
您好!
作为副注,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 loadDllFunc
给出正确的类型:
std :: function< T> func(reinterpret_cast< __ stdcall T *>(lpfnGetProcessID));
似乎将其强制转换为 __ stdcall int(*)(int)
应该是 int(__stdcall *)(int)
。
这里是另一个使用辅助解析器类实现 loadDllFunc
的方法。此解决方案将正确将函数指针转换为 int(__stdcall *)(int)
。
模板< typename T>
struct TypeParser {};
模板< 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));
}
};
模板< typename T>
std :: function< T> loadDllFunc(const std :: string& dllName,const std :: string& funcName){
//加载DLL。
HINSTANCE hGetProcIDDLL = LoadLibrary(dllName.c_str());
//检查DLL是否已加载。
if(hGetProcIDDLL == NULL){
std :: cerr<< 无法加载DLL \<< dllName<< \<< std :: endl;
出口(EXIT_FAILURE);
}
//在DLL中定位函数。
FARPROC lpfnGetProcessID = GetProcAddress(hGetProcIDDLL,funcName.c_str());
//检查是否找到函数。
if(!lpfnGetProcessID){
std :: cerr<< 无法在DLL中找到函数\ funcName << \ \ <<< dllName << \ < std :: endl;
出口(EXIT_FAILURE);
}
//从函数指针创建函数对象。
返回TypeParser< T> :: createFunction(lpfnGetProcessID);
}
How can I load an arbitrary dynamic-link library (dll) function into a std::function
object using a single function?
For example I would like to compile two functions into a dll:
// test.dll
int plusFive(int value) {
return value + 5;
}
void printHello() {
std::cout << "Hello!" << std::endl;
}
And load them both at runtime using a single function like this:
// main.cc
#include <functional>
int main() {
std::function<int(int)> func1(loadDllFunc("test.dll", "plusFive"));
std::function<void()> func2(loadDllFunc("test.dll", "printHello"));
}
Use the WinAPI functions provided in windows.h
(descriptions taken from MSDN Dev Center).
LoadLibrary
- Loads the specified module into the address space of the calling process. Returns a handle to the module.GetProcAddress
- Retrieves the address of an exported function or variable from the specified dynamic-link library (DLL). Returns the address of the exported function or variable.
Use this function to load a specific function and return a std::function
object:
// 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;
}
The DLL source should be written like this:
// 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;
}
And then use loadDllFunc
like this:
// 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();
}
Output:
Result of func1: 6
Hello!
As a sidenote the DLL can be compiled using GCC (4.7.2) like this:
g++ -shared -o test.dll test.cc -std=c++11
Edit:
I'm not sure that the cast in loadDllFunc
gives the correct type:
std::function<T> func(reinterpret_cast<__stdcall T*>(lpfnGetProcessID));
It seems to cast it to __stdcall int (*)(int)
when it should be int (__stdcall *)(int)
.
Here is another way to implement loadDllFunc
using an auxiliary parser class. This solution will correctly cast the function pointer to 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);
}
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