c++ - 如何在 C++20 的模板化仿函数中处理 void 返回类型？

Question

我已经构建了一个模板化仿函数对象，我可以使用它来管理需要递归和跨范围生存的 lambda。它不是很漂亮（它使用 void 指针和一个std::function实例），但它适用于我需要它的用例。（如果张贴者可以保留评论它是如何不安全且非常糟糕的做法，我将不胜感激。我知道。）

但是，它有一个明显的问题：它无法处理 returnvoid的 lambda，因为某些路径试图将返回值存储在变量中。我需要知道如何使用if constexpr语句来检测仿函数的 lambda 的结果是否为 void，并适当地处理它。这不是一个独特的问题，但我发现的所有结果都非常过时，其中许多都使用现已折旧的result_of_t.

任何帮助将不胜感激。

---

#include <iostream>
#include <string>
#include <functional>

#define uint unsigned int

//! A standardised wrapper for lambda functions, which can be stored in pointers, used recursively, keep track of external storage via a void *, and set to self destruct when no longer useful.
template <class F, bool UsesDataStorage>
class Functor {
protected:
    std::function<F> m_f; //!< The lambda stored by the wrapper
    void* m_data = nullptr; //!< A void pointer which will be given to the lambda if `UsesDataStorage`. Note that cleanup is delegated to the lambda; the functor instance will not handle it.
    bool m_selfDestructing = true; //!< Whether the combinator will self-destruct should its lambda mark itself as no longer useful.
    bool m_selfDestructTrigger = false; //!< Whether the combinator's lambda has marked itself as no longer useful.
public:
    inline bool usesDataStorage() const { return UsesDataStorage; } //!< Return whether this functor is set up to give its function a `data` void-pointer, which will presumably be set to a data-structure.
    inline void* getData() const { return m_data; } //!< Returns the void pointer which is passed to the lambda at each call (if the functor instance uses data storage).
    inline void setData(void* data) { m_data = data; }  //!< Sets the void pointer which is passed to the lambda at each call (if the functor instance uses data storage).
    inline bool canSelfDestruct() const { return m_selfDestructing; } //!< Returns whether the LambdaWrapper will delete itself when instructed to by the contained lambda.
    inline void triggerSelfDestruct() { m_selfDestructTrigger = true; } //!< Triggers wrapper self-deletion at the end of ruinning the lambda.

    Functor(const std::function<F>& f, bool canSelfDestruct = true) :
        m_f(f),
        m_selfDestructing(canSelfDestruct)
    {} //!< Constructor for Functor instances which DON'T use data storage. Note that the given function should always take a void pointer as the first argument, which is where a pointer to the Functor instance will be passed.
    Functor(std::function<F>&& f, bool canSelfDestruct = true) :
        m_f(f),
        m_selfDestructing(canSelfDestruct)
    {} //!< Constructor for Functor instances which DON'T use data storage. Note that the given function should always take a void pointer as the first argument, which is where a pointer to the Functor instance will be passed.
    Functor(const std::function<F>& f, void* data, bool canSelfDestruct = true) :
        m_f(f),
        m_data(data),
        m_selfDestructing(canSelfDestruct)
    {} //!< Constructor for Functor instances which DO use data storage. Note that the given function should always take a void pointer as the first argument, which is where a pointer to the Functor instance will be passed, and a void * for the second argument, which is where the data storage pointer is passed.
    Functor(std::function<F>&& f, void* data, bool canSelfDestruct = true) :
        m_f(f),
        m_data(data),
        m_selfDestructing(canSelfDestruct)
    {} //!< Constructor for Functor instances which DO use data storage. Note that the given function should always take a void pointer as the first argument, which is where a pointer to the Functor instance will be passed, which is where the data storage pointer is passed.

    template <typename... Args>
    decltype(auto) operator()(Args&&... args) {
        // Avoid storing return if we can, 
        if (!m_selfDestructing) {
            if constexpr (UsesDataStorage) {
                // Pass itself to m_f, then the data storage, then the arguments.
                // This should work even if the return type is void, as far as I can tell.
                return m_f(this, m_data, std::forward<Args>(args)...);
            }
            else {
                // Pass itself to m_f, then the arguments.
                // This should work even if the return type is void, as far as I can tell.
                return m_f(this, std::forward<Args>(args)...);
            }
        }
        else {
            if constexpr (UsesDataStorage) {
                // Pass itself to m_f, then the data storage, then the arguments.

                // ----- !!! -----
                // The following if constexpr statement is what I can't work out how to do.
                // ----- !!! -----
                if constexpr (std::is_same<std::invoke_result_t<std::function<F>>, void>) {
                    m_f(this, m_data, std::forward<Args>(args)...);
                    // self-destruct if necessary, allowing lamdas to delete themselves if they know they're no longer useful.
                    if (m_selfDestructTrigger) { delete this; }
                    return;
                }
                else {
                    auto r = m_f(this, m_data, std::forward<Args>(args)...);
                    // self-destruct if necessary, allowing lamdas to delete themselves if they know they're no longer useful.
                    if (m_selfDestructTrigger) { delete this; }
                    return r;
                }
            }
            else {
                // Pass itself to m_f, then the arguments.

                // ----- !!! -----
                // The following if constexpr statement is what I can't work out how to do.
                // ----- !!! -----
                if constexpr (std::is_same<std::invoke_result_t<std::function<F>>, void>) {
                    m_f(this, std::forward<Args>(args)...);
                    // self-destruct if necessary, allowing lamdas to delete themselves if they know they're no longer useful.
                    if (m_selfDestructTrigger) { delete this; }
                    return;
                }
                else {
                    auto r = m_f(this, std::forward<Args>(args)...);
                    // self-destruct if necessary, allowing lamdas to delete themselves if they know they're no longer useful.
                    if (m_selfDestructTrigger) { delete this; }
                    return r;
                }
            }
        }
    }
};
template <class F> Functor(std::function<F>, bool)->Functor<F, false>;
template <class F> Functor(std::function<F>, void*, bool)->Functor<F, true>;

int main() {
    Functor f1 = Functor(std::function([](void* self, uint val1) -> uint {
        std::cout << "f1(" << val1 << ") was called." << std::endl;
        return 2u * val1;
    }), false);
    Functor f2 = Functor(std::function([](void* self, uint val1) -> void {
        std::cout << "f2(" << val1 << ") was called." << std::endl;
        return;
    }), false);

    auto x = f1(3u); // Compiles and works.
    f2(3u); // Doesn't compile.
}

标签： c++templatesc++20functor

Answer 1

我正在寻找的行是这样的：

if constexpr (std::is_same<std::function<F>::result_type, void>::value) {}

非常感谢@NathanOliver！