使函数接受可选选项以接受非可选选项吗? [英] Make a function accepting an optional to accept a non-optional?

问题描述

我正在尝试在std::optional上以monad风格编写语法糖.请考虑:

I'm trying to write syntactic sugar, in a monad-style, over std::optional. Please consider:

template<class T>
void f(std::optional<T>)
{}

就这样，即使存在从T到std::optional<T> <的转换，也不能使用非可选的T ¹ (例如int)调用此函数. sup> 2 .

As is, this function cannot be called with a non-optional T¹ (e.g. an int), even though there exists a conversion from T to std::optional<T>².

是否有一种方法可以使f接受std::optional<T>或T(在调用者站点上转换为可选)，而无需定义重载 ³ ?

Is there a way to make f accept an std::optional<T> or a T (converted to an optional at the caller site), without defining an overload³?

¹⁾ f(0):error: no matching function for call to 'f(int)'和note: template argument deduction/substitution failed，(

¹⁾ f(0): error: no matching function for call to 'f(int)' and note: template argument deduction/substitution failed, (demo).
²⁾ Because template argument deduction doesn't consider conversions.
³⁾ Overloading is an acceptable solution for a unary function, but starts to be an annoyance when you have binary functions like operator+(optional, optional), and is a pain for ternary, 4-ary, etc. functions.

推荐答案

另一个版本.这不涉及任何内容:

Another version. This one doesn't involve anything:

template <typename T>
void f(T&& t) {
    std::optional opt = std::forward<T>(t);
}

类模板参数推论已经在这里做对了.如果t是optional，将首选复制扣除对象，并且我们会得到相同的类型.否则，我们将其包装.

Class template argument deduction already does the right thing here. If t is an optional, the copy deduction candidate will be preferred and we get the same type back. Otherwise, we wrap it.

这篇关于使函数接受可选选项以接受非可选选项吗?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！