constexpr和RTTI [英] constexpr and RTTI
问题描述
我想做这样的事情:
template <typename T>
constexpr ::std::size_t type_name_hash()
{
return ::std::hash<::std::string>()(typeid(T).name());
}
现在,我都不知道 hash
或 string
都是 constexpr
,但这可以解决,假设它们是 constexpr
。我想问的是,如果打开 RTTI
,是否应该使用 constexpr
函数计算<$ c的哈希值$ c> typeid(T).name()仍然会产生一个编译时常量吗?如何关闭 RTTI
?
Now, I know neither hash
nor string
are constexpr
, but this could be worked around, assume they are constexpr
. What I want to ask is, if RTTI
was turned on, should a constexpr
function computing a hash of typeid(T).name()
still produce a compile-time constant? How about when RTTI
is turned off?
推荐答案
运行时类型标识,您认为在编译时可行吗?常量表达式的规则不允许:
What part of Run-Time Type Identification do you think works at compile-time? The rules for constant expressions disallow:
-一个
typeid
表达式(5.2.8 ),其操作数是多态类类型的glvalue;
— a
typeid
expression (5.2.8) whose operand is a glvalue of a polymorphic class type;
,因此您的模板仅适用于某些类型。
so your template would only work for some types.
并且在关闭RTTI的情况下,您根本不能使用 typeid
。
And with RTTI turned off you can't use typeid
at all.
C ++ 11已经提供了一种哈希类型的机制:
C++11 already provides a mechanism to hash a type:
return ::std::hash<::std::type_index>()(::std::type_index(typeid(T)));
但这并不是所有类型的常量表达式。
But it's not going to be a constant expression for all types.
您可以使用指向每种类型的指针的 type_index
,因为指针不是多态类类型,并且仍会给出唯一的类型:
You could use the type_index
of a pointer to each type, as a pointer is not a polymorphic class type and will still give a unique type:
return ::std::hash<::std::type_index>()(::std::type_index(typeid(T*)));
现在的问题是仅是 type_index
构造函数不是 constexpr
,散列函数也不是。
Now the problem is "only" that the type_index
constructor is not constexpr
and neither is the hash function.
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