在C ++中编译时间类型确定 [英] Compile time type determination in C++
问题描述
同事最近向我展示了一些他在网上找到的代码。它似乎允许编译时确定类型是否具有与另一个类型的is a关系。我认为这是完全真棒,但我必须承认,我没有这个实际工作原理。
A coworker recently showed me some code that he found online. It appears to allow compile time determination of whether a type has an "is a" relationship with another type. I think this is totally awesome, but I have to admit that I'm clueless as to how this actually works. Can anyone explain this to me?
template<typename BaseT, typename DerivedT>
inline bool isRelated(const DerivedT&)
{
DerivedT derived();
char test(const BaseT&); // sizeof(test()) == sizeof(char)
char (&test(...))[2]; // sizeof(test()) == sizeof(char[2])
struct conversion
{
enum { exists = (sizeof(test(derived())) == sizeof(char)) };
};
return conversion::exists;
}
$ b >
Once this function is defined, you can use it like this:
#include <iostream>
class base {};
class derived : public base {};
class unrelated {};
int main()
{
base b;
derived d;
unrelated u;
if( isRelated<base>( b ) )
std::cout << "b is related to base" << std::endl;
if( isRelated<base>( d ) )
std::cout << "d is related to base" << std::endl;
if( !isRelated<base>( u ) )
std::cout << "u is not related to base" << std::endl;
}
推荐答案
它声明两个重载的函数 test
,一个采用 Base
,一个采取任何 c $ c>,并返回不同的类型。
It declares two overloaded functions named test
, one taking a Base
and one taking anything (...)
, and returning different types.
然后调用 Derived
的返回类型来查看调用了哪个重载。 (它实际上调用函数的返回值返回 Derived
,以避免使用内存)
It then calls the function with a Derived
and checks the size of its return type to see which overload is called. (It actually calls the function with the return value of a function that returns Derived
, to avoid using memory)
因为 enum
是编译时常量,所有这些都是在编译时在类型系统中完成的。由于函数不会在运行时被调用,因此它们没有主体没有关系。
Because enum
s are compile-time constants, all of this is done within the type system at compile-time. Since the functions don't end up getting called at runtime, it doesn't matter that they have no bodies.
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