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> T (例如int)调用此函数. sup> 2 .
As is, this function cannot be called with a non-optional T (e.g. an int), even though there exists a conversion from T to std::optional<T>.
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 ?
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 overload?
f(0):error: no matching function for call to 'f(int)'和note: template argument deduction/substitution failed，(演示).
因为模板参数推导不考虑转换.
对于一元函数，重载是可以接受的解决方案，但是当您拥有像operator+(optional, optional)这样的二进制函数时，重载就变得很烦人，而对于三元，4元，等而言，这是一个痛苦. 函数.
f(0): error: no matching function for call to 'f(int)' and note: template argument deduction/substitution failed, (demo).
Because template argument deduction doesn't consider conversions.
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.
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