C++ template function with std::optional [duplicate] - c++

Question

I'm trying to write syntactic sugar, in a monad-style, over std::optional. Please consider:
template<class T>
void f(std::optional<T>)
{}
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.
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)' 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.

Answer 1

Another version. This one doesn't involve anything:
template <typename T>
void f(T&& t) {
std::optional opt = std::forward<T>(t);
}
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.

Answer 2

Instead of taking optional as argument take deductible template parameter:
template<class T>
struct is_optional : std::false_type{};
template<class T>
struct is_optional<std::optional<T>> : std::true_type{};
template<class T, class = std::enable_if_t<is_optional<std::decay_t<T>>::value>>
constexpr decltype(auto) to_optional(T &&val){
return std::forward<T>(val);
}
template<class T, class = std::enable_if_t<!is_optional<std::decay_t<T>>::value>>
constexpr std::optional<std::decay_t<T>> to_optional(T &&val){
return { std::forward<T>(val) };
}
template<class T>
void f(T &&t){
auto opt = to_optional(std::forward<T>(t));
}
int main() {
f(1);
f(std::optional<int>(1));
}
Live example

Answer 3

This uses one of my favorite type traits, which can check any all-type template against a type to see if it's the template for it.
#include <iostream>
#include <type_traits>
#include <optional>
template<template<class...> class tmpl, typename T>
struct x_is_template_for : public std::false_type {};
template<template<class...> class tmpl, class... Args>
struct x_is_template_for<tmpl, tmpl<Args...>> : public std::true_type {};
template<template<class...> class tmpl, typename... Ts>
using is_template_for = std::conjunction<x_is_template_for<tmpl, std::decay_t<Ts>>...>;
template<template<class...> class tmpl, typename... Ts>
constexpr bool is_template_for_v = is_template_for<tmpl, Ts...>::value;
template <typename T>
void f(T && t) {
auto optional_t = [&]{
if constexpr (is_template_for_v<std::optional, T>) {
return t;
} else {
return std::optional<std::remove_reference_t<T>>(std::forward<T>(t));
}
}();
(void)optional_t;
}
int main() {
int i = 5;
std::optional<int> oi{5};
f(i);
f(oi);
}
https://godbolt.org/z/HXgoEE

Answer 4

Another version. This one doesn't involve writing traits:
template <typename T>
struct make_optional_t {
template <typename U>
auto operator()(U&& u) const {
return std::optional<T>(std::forward<U>(u));
}
};
template <typename T>
struct make_optional_t<std::optional<T>> {
template <typename U>
auto operator()(U&& u) const {
return std::forward<U>(u);
}
};
template <typename T>
inline make_optional_t<std::decay_t<T>> make_optional;
template <typename T>
void f(T&& t){
auto opt = make_optional<T>(std::forward<T>(t));
}