C ++试图从std ::函数获取函数地址 [英] C++ trying to get function address from a std::function

问题描述

我试图从std ::函数中找到函数的地址。

i'm trying to find the address of a function from a std::function.

第一个解决方案是：

size_t getAddress(std::function<void (void)> function) {
    typedef void (fnType)(void);
    fnType ** fnPointer = function.target<fnType *>();
    return (size_t) *fnPointer;
}

但是这只适用于带有需要的功能
签名是（void（Type&）），我试图做

But that only works for function with (void ()) signature, since i need for function that signature are (void (Type &)), i tried to do

template<typename T>
size_t getAddress(std::function<void (T &)> function) {
    typedef void (fnType)(T &);
    fnType ** fnPointer = function.target<fnType *>();
    return (size_t) *fnPointer;
}

我得到Error - expected类型构造

And i get "Error - expected '(' for function-style cast or type construction"

更新：是否有任何方法捕获成员类地址？对类成员我使用：

Update: Is any way to capture member class address? for class members i'm using:

template<typename Clazz, typename Return, typename ...Arguments>
size_t getMemberAddress(std::function<Return (Clazz::*)(Arguments...)> & executor) {
    typedef Return (Clazz::*fnType)(Arguments...);
    fnType ** fnPointer = executor.template target<fnType *>();
    if (fnPointer != nullptr) {
        return (size_t) * fnPointer;
    }
    return 0;
}

更新：捕获lambda我正在使用

Update: To capture lambda i'm using

template <typename Function>
struct function_traits
    : public function_traits<decltype(&Function::operator())> {
};

template <typename ClassType, typename ReturnType, typename... Args>
struct function_traits<ReturnType(ClassType::*)(Args...) const> {
    typedef ReturnType (*pointer)(Args...);
    typedef std::function<ReturnType(Args...)> function;
};

template <typename Function>
typename function_traits<Function>::function
to_function (Function & lambda) {
    return static_cast<typename function_traits<Function>::function>(lambda);
}

template <typename Lambda>
size_t getAddress(Lambda lambda) {
    auto function = new decltype(to_function(lambda))(to_function(lambda));
    void * func = static_cast<void *>(function);
    return (size_t)func;
}

std::cout << getAddress([] { std::cout << "Hello" << std::endl;}) << std::endl;

推荐答案

您需要使用模板关键字：

#include <functional>
#include <iostream>

template<typename T>
size_t getAddress(std::function<void (T &)> f) {
    typedef void (fnType)(T &);
    fnType ** fnPointer = f.template target<fnType*>();
    return (size_t) *fnPointer;
}

void foo(int& a) {
    a = 0;
}

int main() {
    std::function<void(int&)> f = &foo;
    std::cout << (size_t)&foo << std::endl << getAddress(f) << std::endl;
    return 0;
}

提示：当您遇到C ++语法问题时，建议您使用 clang ++ 来编译你的代码。如果你玩弄你的编写代码，通常指向你的写入方向来修正错误（当它可以找出你在做什么）。

Hint: When you have problems with C++ syntax, I suggest you use clang++ to compile your code. If you play around with how your write the code it will usually point you in the write direction to fix the error (when it can figure out what you are doing).

我还建议您使用可变参数模板，使您的函数更加一般：

I also suggest that you use variadic templates to make your function a bit more general:

template<typename T, typename... U>
size_t getAddress(std::function<T(U...)> f) {
    typedef T(fnType)(U...);
    fnType ** fnPointer = f.template target<fnType*>();
    return (size_t) *fnPointer;
}

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