是否有间接函子? [英] Is There a Indirection Functor?

问题描述

我正在寻找一元函子，该函子将取消引用它的参数并返回结果.当然，我可以写一个，似乎应该已经存在了.

I'm looking for a unary functor which will dereference it's argument and return the result. Of course I can write one, it just seemed like something should already exist.

因此给出了代码:

const auto vals = { 0, 1, 2, 3 };
vector<const int*> test(size(vals), nullptr);

iota(begin(test), end(test), data(vals));

transform(cbegin(test), cend(test), ostream_iterator<int>(cout, " "), [](const auto& i){ return *i; });

实时示例

我希望有一个可以代替lambda的仿函数.这样的事情是否存在，还是我只需要使用lambda?

I was hoping that there was a functor that I could use instead of the lambda. Does such a thing exist, or do I need to just use the lambda?

推荐答案

假定"functor" 是指功能对象" 或可调用对象"，标准库中似乎没有您想要的东西.

自己实现它很简单:

struct deferencer
{
    template <typename T>
    decltype(auto) operator()(T&& x) const
        noexcept(noexcept(*x))
    { 
        return *x; 
    }
};

---

请注意，您的lambda并没有达到您的期望，因为它的隐式返回类型为-> auto，它会进行复制.一种可能的正确lambda是:

---

Note that your lambda doesn't do what you expect, as its implicit return type is -> auto, which makes a copy. One possible correct lambda is:

[](const auto& i) -> decltype(auto) { return *i; }

如果您没有为lambda指定显式的跟踪返回类型，则隐式的将是auto，它始终是副本. operator*返回引用无关紧要，因为lambda返回副本(即operator*返回的引用然后由lambda的return语句复制).

If you don't specify an explicit trailing return type for a lambda, the implicit one will be auto which is always a copy. It doesn't matter if operator* returns a reference, since the lambda returns a copy (i.e. the reference returned by operator* is then copied by the lambda's return statement).

struct A
{
    A() = default;
    A(const A&) { puts("copy ctor\n"); }
};

int main()
{
    []{ return *(new A); }(); // prints "copy ctor"
}

wandbox示例

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