Files
callable_traits/example/is_constexpr_function_object.cpp
2016-04-15 21:41:42 -05:00

95 lines
2.4 KiB
C++

/*<-
Copyright Barrett Adair 2016
Distributed under the Boost Software License, Version 1.0.
(See accompanying file LICENSE.md or copy at http ://boost.org/LICENSE_1_0.txt)
->*/
#include <callable_traits/config.hpp>
#ifdef CALLABLE_TRAITS_DISABLE_CONSTEXPR_CHECKS
int main(){ return 0; }
#else
//[ is_constexpr_function_object
/*`[warning When compiling in MSVC, `is_constexpr` always returns
`std::false_type`, because MSVC cannot compile the logic that normally determines
this.]*/
#include <type_traits>
#include <callable_traits/is_constexpr.hpp>
namespace ct = callable_traits;
//this is a constexpr function object (non-templated)
struct zero {
constexpr auto operator()() const {
return 0;
}
};
static_assert(ct::is_constexpr<zero>(), "");
static_assert(ct::is_constexpr(zero{}), "");
//this is a constexpr function object (templated)
struct subtract {
// For callable_traits::is_constexpr, generic function objects
// only need to be SFINAE-friendly if the body of the operator()
// function accesses member names besides "type" and "value".
// Unary/binary operators and constructor calls are okay to use.
template<typename T1, typename T2>
constexpr auto operator()(T1, T2) const {
return T1{} - T2{};
}
};
static_assert(ct::is_constexpr<subtract>(), "");
static_assert(ct::is_constexpr(subtract{}), "");
//this is NOT a constexpr function object
struct add {
template<typename T1, typename T2>
auto operator()(T1, T2) const {
return T1{} + T2{};
}
};
static_assert(!ct::is_constexpr<add>(), "");
static_assert(!ct::is_constexpr(add{}), "");
auto multiply = [](auto t1, auto t2) -> decltype(t1.value * t2.value) {
return t1.value * t2.value;
};
static_assert(!ct::is_constexpr<decltype(multiply)>(), "");
static_assert(!ct::is_constexpr(multiply), "");
// is_constexpr will always return std::false_type when the argument
// is either not a literal type, or is not default constructible. Below,
// divide is not default constructible, so is_constexpr returns
// std::false_type. For literal types that are default constructible, a
// constexpr default constructor is assumed.
struct divide {
divide() = delete;
constexpr divide(int){};
template<typename T1, typename T2>
constexpr auto operator()(T1, T2) const {
return T1{} / T2{};
}
};
static_assert(!ct::is_constexpr<divide>(), "");
static_assert(!ct::is_constexpr(divide{0}), "");
int main() {}
//]
#endif