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I can't find a nice way to define generic higher-order functions taking generic functions as arguments. For example, take this attempt at one of the simplest such functions out there:

template<typename F, typename A>
auto apply(F f, const A& a) -> decltype(f(a)){return f(a);}

Of course it works as intended when used with non-template functions. But if I have, for instance

template<typename A>
A id(const A& a){return a;}

then

int a = 10;
int b = apply(id, a);

will not work, since id expects a template parameter. I could just write id<int> to make it work, but that sort of defeats the purpose (as I see it, it implies that if I wanted to write "filter" I'd have to write a separate definition for each generic predicate function). Using std::function or function pointers did not help. Also, I tried to make a "template template" version of apply, but I get various compiler errors when I try to use it:

template<template<typename> class F, typename A>
auto apply2(F<A> f, const A& a)-> decltype(f(a)){return f(a);}

The best I came up with was the following:

struct Id{
    template<typename A>
    static A func(const A& a){return a;}
};

template<typename F, typename A>
auto apply(A a)-> decltype(F::func(a)){return F::func(a);}

It's a bit ugly, but now at least I can actually parameterize by the function.

So, is there a better way to do generic functions taking generic functions as arguments?

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2 回答 2

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This 'functor' works with your first version of apply.

struct id
{
    template<typename A>
    A operator ()(const A& a) {return a;}
};

later

int b = apply(id(), a); // with extra parenthesis to construct the struct.
于 2013-09-04T15:57:05.933 回答
1

在 C++11 中，您可以做的最好的事情是使用struct带有模板化函数调用运算符的 a：

struct ID {
    template <typename T>
    void operator()(T arg) const {
        ...
    }
};

这比函数 [pointer] 更可取的原因是它更有可能被内联。使用 C++14，您可以使用 lambda：

[](auto arg){ ... }

这只是写struct上面的一个可爱的版本。

于 2013-09-04T16:04:13.490 回答