我已经构建了一个模板化的函数对象，我可以用它来管理需要递归和跨作用域生存的lambda。它不是很漂亮(它使用void指针和std::function实例)，但它适用于我需要它的用例。(如果发帖者可以保留关于它是如何不类型安全的和非常糟糕的做法的评论，我会很感激。我知道。)
然而，有一个明显的问题:它不能处理返回void的lambdas，因为一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语句来检测函子的lambda的结果是否为空，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，其中许多使用现在贬值的result_of_t。
任何帮助都将是非常感激的。
#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.
}
我要找的行是这样的:
if constexpr (std::is_same<std::function<F>::result_type, void>::value) {}