在函数对象中包装模板或重载函数的最简洁和可重用的方法 [英] Most terse and reusable way of wrapping template or overloaded functions in function objects

问题描述

场景一:模板函数pred

template<typename T>
bool pred(T t) { /* return a bool based on t */ }

场景二:一组重载同名的函数pred

Scenario 2: a set of functions overloaded on the same name pred

bool pred(A t) { /* return a bool based on t */ }
bool pred(B t) { /* return a bool based on t */ }
bool pred(C t) { /* return a bool based on t */ }
...

无论我们处于这两种情况中的哪一种，最重要的是 pred 不引用函数，因此它不能被传递，例如作为 std::remove_if 的一元谓词.

Whichever of the two scenarii we're in, the bottom line is that pred does not refer to a function, and so it cannot be passed around, e.g. as a unary predicate to std::remove_if.

因此在这种情况下定义以下可以传递的对象很方便，

Therefore it is convenient in this case to define the following object which can be passed around instead,

auto constexpr predObj = [](auto t){ return pred(t); };

但是，一旦我对另一个一元谓词有类似的需求，我需要复制并粘贴该行并将两个名称更改为其他名称；同样，如果我需要为二元谓词这样做:

However, as soon as I have a similar need for another unary predicate, I need to copy and paste that line and change the two names to something else; similarly if I need to do that for a binary predicate:

auto contexpr binPredObj = [](auto x, auto y){ return binPred(x, y); };

是否有自动制作的通用方法?我正在考虑类似的事情

Is there a general way of making this automatically? I'm thinking of something like

auto funObj = fun2Obj(fun);

我觉得我问的是不可能的，因为它需要传递 fun 因为它是一个函数对象，它不是，否则我不需要做一个函数对象出来了.但问永远不会犯罪，对吧?

I have the feeling that what I ask is not possible exactly because it would require passing fun as it was a function object, which it isn't, otherwise I wouldn't need to make a function object out of it. But asking is never crime, right?

推荐答案

你可以像这样创建宏

#define FUNCTORIZE(func) [](auto&&... val) \
noexcept(noexcept(func(std::forward<decltype(val)>(val)...))) -> decltype(auto) \
{return func(std::forward<decltype(val)>(val)...);}

它将让您将任何可调用对象包装到一个闭包对象中.你会像这样使用它

which will let you wrap any callable into a closure object. You would use it like

auto constexpr predObj = FUNCTORIZE(pred);

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