C ++检查语句是否可以评估为constexpr [英] C++ check if statement can be evaluated constexpr
问题描述
是否有一种方法可以确定是否可以对constexpr求值,并将结果用作constexpr布尔值?我的简化用例如下:
Is there a method to decide whether something can be constexpr evaluated, and use the result as a constexpr boolean? My simplified use case is as follows:
template <typename base>
class derived
{
template<size_t size>
void do_stuff() { (...) }
void do_stuff(size_t size) { (...) }
public:
void execute()
{
if constexpr(is_constexpr(base::get_data())
{
do_stuff<base::get_data()>();
}
else
{
do_stuff(base::get_data());
}
}
}
我的目标是C ++ 2a。
My target is C++2a.
我找到了以下reddit线程,但我而不是宏的忠实支持者。 https://www.reddit.com/ r / cpp / comments / 7c208c / is_constexpr_a_macro_that_check_if_an_expression /
I found the following reddit thread, but I'm not a big fan of the macros. https://www.reddit.com/r/cpp/comments/7c208c/is_constexpr_a_macro_that_check_if_an_expression/
推荐答案
这是另一种解决方案,它更通用(适用于任何表达式,而无需每次都定义单独的模板。)
Here's another solution, which is more generic (applicable to any expression, without defining a separate template each time).
此解决方案利用了(1)lambda表达式可以是从C ++ 17开始的constexpr(2)从C ++ 20开始默认可构造的lambda类型。
This solution leverages that (1) lambda expressions can be constexpr as of C++17 (2) the type of a captureless lambda is default constructible as of C++20.
这个想法是,返回 true
仅当且仅当 Lambda {}()
可以出现在模板参数中时才选择,这实际上需要lambda调用
The idea is, the overload that returns true
is selected when and only when Lambda{}()
can appear within a template argument, which effectively requires the lambda invocation to be a constant expression.
template<class Lambda, int=(Lambda{}(), 0)>
constexpr bool is_constexpr(Lambda) { return true; }
constexpr bool is_constexpr(...) { return false; }
template <typename base>
class derived
{
// ...
void execute()
{
if constexpr(is_constexpr([]{ base::get_data(); }))
do_stuff<base::get_data()>();
else
do_stuff(base::get_data());
}
}
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