使可变函数接受任意函子并返回输入函子的每个返回值的元组 [英] Make variadic function which takes arbitary functors and returns a tuple of each return value of input functors

问题描述

我想创建一个函数对象，它接受任意函数对象，并返回一个元组，它存储每个函数对象的返回值。

I want to make a function object which takes arbitrary function objects and returns a tuple which stores the return value of each function object.

为了达到这个目标，我做了一个 A类

To achieve this goal, I made a class A

class A
{
private:
    template <class Ret, class Func>
    auto impl(Ret ret, Func func) -> decltype(tuple_cat(ret, make_tuple(func())))
    {
        return tuple_cat(ret, make_tuple(func()));
    }

    template <class Ret, class First, class... Funcs>
    auto impl(Ret ret, First first, Funcs... funcs) 
    -> decltype(impl(tuple_cat(ret, make_tuple(first())), funcs...))
    {
    return impl(tuple_cat(ret, make_tuple(first())), funcs...);
    }

public:
    template <class Func>
    auto operator()(Func func) -> decltype(make_tuple(func()))
        {
        return make_tuple(func());
    }

    template <class First, class... Funcs>
    auto operator()(First first, Funcs... funcs)
     -> decltype(impl(make_tuple(first()),funcs...))
    {
        impl(make_tuple(first()),funcs...);
    }
};

在主要功能中，我做了三个羔羊。

And in the main function, I made three lambdas.

int main(){
    auto func1 = [](){ cout << 1 << endl; return 1;};
    auto func2 = [](){ cout << 2 << endl; return 2;};
    auto func3 = [](){ cout << 3 << endl; return 3;};

    A a;
    auto x = a(func1, func2);
    cout << "ans : " << get<0>(x) << get<1>(x) << endl; // I expect ans : 12
}

这段代码可以通过gcc 4.7编译。 2。但是，它不工作，因为我的预期。
我应该如何修改此代码？

This code can be compiled by gcc 4.7.2. However, it doesn't work as I expected. How should I modify this code?

推荐答案

我认为问题是你缺少一个 return 语句：

I think the problem is that you're missing a return statement:

template <class First, class... Funcs>
auto operator()(First first, Funcs... funcs)
 -> decltype(impl(make_tuple(first()),funcs...))
{
    return impl(make_tuple(first()),funcs...);
//  ^^^^^^
}

代码具有未定义的行为。根据C ++ 11标准的第6.6.3 / 2节：

Without it, your code has Undefined Behavior. Per Paragraph 6.6.3/2 of the C++11 Standard:

[...]相当于没有价值的回报;这会导致价值回报函数中的未定义行为。

[...] Flowing off the end of a function is equivalent to a return with no value; this results in undefined behavior in a value-returning function.

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