更新时间:2023-11-27 18:41:28
#define HasMember(NAME)
template<class Class, typename Type = void>
struct HasMember_##NAME
{
typedef char (&yes)[2];
template<unsigned long> struct exists;
template<typename V> static yes Check (exists<sizeof(static_cast<Type>(&V::NAME))>*);
template<typename> static char Check (...);
static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes));
};
template<class Class>
struct HasMember_##NAME<Class, void>
{
typedef char (&yes)[2];
template<unsigned long> struct exists;
template<typename V> static yes Check (exists<sizeof(&V::NAME)>*);
template<typename> static char Check (...);
static const bool value = (sizeof(Check<Class>(0)) == sizeof(yes));
}
用法:只需使用您想要查找的任何成员调用宏:
Usage: Simply invoke the macro with whatever member you want to find:
HasMember(Foo); // Creates a SFINAE `class HasMember_Foo`
HasMember(i); // Creates a SFINAE `class HasMember_i`
现在我们可以使用HasMember_X
来检查任何class
中的X
,如下所示:
Now we can utilize HasMember_X
to check X
in ANY class
as below:
#include<iostream>
struct S
{
void Foo () const {}
// void Foo () {} // If uncommented then type should be mentioned in `HasMember_Foo`
int i;
};
int main ()
{
std::cout << HasMember_Foo<S, void (S::*) () const>::value << "
";
std::cout << HasMember_Foo<S>::value << "
";
std::cout << HasMember_i<S, int (S::*)>::value << "
";
std::cout << HasMember_i<S>::value << "
";
}
捕获:
class
不得有重载方法.如果有,那么这个技巧就失败了.即,即使命名成员出现不止一次,结果也会是 false
.B
是 S
的基础 &void B::Bar()
存在,则 HasMember_Bar::value
或 HasMember_Bar::value
或 HasMember_Bar::value
将给出 false
class
must not have overloaded methods. If it has then this trick fails.
i.e. even though the named member is present more than once, the result will be false
.B
is base of S
& void B::Bar ()
is present, then HasMember_Bar<S, void (B::*)()>::value
or HasMember_Bar<S, void (S::*)()>::value
or HasMember_Bar<S>::value
will give false