:: std :: function的constexpr版本 [英] constexpr version of ::std::function

问题描述

我正在寻找可在constexpr中使用的:: std :: function. 用例:我有一个函数，该函数以函数指针作为参数，第二个函数将lambda传递给第一个函数.两者在编译时都是完全可执行的，因此我想对其进行constexpr. 例如:

I am in search of a ::std::function usable in constexpr. Use case: I have a function which takes a function pointer as an argument, and a second which passes a lambda to the first function. Both are fully executable at compile time, so I want to constexpr them. Eg:

template <class _Type>
class ConstexprFunctionPtr
{
    private:
        using Type = typename ::std::decay<_Type>::type;
        const Type function;

    public:
        constexpr inline
        ConstexprFunctionPtr(const Type f)
        : function(f)
        { }

        template <typename... Types>
        constexpr inline
        auto
        operator() (Types... args)
        const {
            return function(args... );
        }
};

constexpr inline
void
test()
{
    ConstexprFunctionPtr<int(int)> test([](int i) -> int {
        return i + 1;
    });
    int i = test(100);

    ConstexprFunctionPtr<int(int)> test2([=](int i) -> int {
        return i + 1;
    });
    i = test2(1000);
}

但是，这仅起作用，因为我将lambda转换为函数指针，并且当然无法捕获lambda，如第二个示例所示.谁能给我一些有关捕获lambda的方法的提示?

However, this only works because I am converting the lambda to a function pointer, and of course fails with capturing lambdas as showed in the second example. Can anyone give me some pointers on how to do that with capturing lambdas?

这将演示用例:

constexpr
void
walkOverObjects(ObjectList d, ConstexprFunctionPtr<void(Object)> fun) {
// for i in d, execute fun
}

constexpr
void
searchObjectX(ObjectList d) {
walkOverObjects(d, /*lambda that searches X*/);
}

谢谢， 杰克

推荐答案

我正在寻找可在constexpr中使用的:: std :: function

I am in search of a ::std::function usable in constexpr

在这里停止.不可能. std::function是一个多态包装函数.无状态Lambda，有状态Lambda，函子，函数指针，函数引用-它们都可以构建有效的std::function，并且可以在运行时更改.因此，使编译时间等效只是浪费时间.

Stop right here. it's impossible. std::function is a polymorphic wrapper function. stateless lambdas, statefull lambdas, functors, function pointers, function references - all of them can build a valid std::function that can change during runtime. so making a compile time equivalent is just a waste of time.

如果只需要编译时通用函数参数，则可以使用模板

If you just want a compile time generic function parameter, you can just use templates

template<class functor_type>
class my_generic_function_consumer_class{

   using decayed_function_type = typename std::decay_t<functor_type>;

   decayed_function_type m_functor;

};

在有问题的代码中，只需接受通用函子，然后使用static_assert对其进行验证:

In your code in question, just accept a generic functor, and validate it using static_assert:

template<class function_type>
constexpr void walkOverObjects(ObjectList d, function_type&& fun) {
    static_assert(std::is_constructible_v<std::function<void(ObjectList), function_type>>,
                  "function_type given to walkOverObjects is invalid.");
}

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