使用 SFINAE,我 can detect给定的类是否具有特定的成员函数。但是,如果我想测试继承的成员函数怎么办?
以下在 VC8 和 GCC4 中不起作用(即检测到 A
有一个成员函数 foo()
,但没有 B
继承一个):
#include <iostream>
template<typename T, typename Sig>
struct has_foo {
template <typename U, U> struct type_check;
template <typename V> static char (& chk(type_check<Sig, &V::foo>*))[1];
template <typename > static char (& chk(...))[2];
static bool const value = (sizeof(chk<T>(0)) == 1);
};
struct A {
void foo();
};
struct B : A {};
int main()
{
using namespace std;
cout << boolalpha << has_foo<A, void (A::*)()>::value << endl; // true
cout << boolalpha << has_foo<B, void (B::*)()>::value << endl; // false
}
那么,有没有办法测试继承的成员函数呢?
最佳答案
看看这个线程:
http://lists.boost.org/boost-users/2009/01/44538.php
源自该讨论中链接的代码:
#include <iostream>
template <typename Type>
class has_foo
{
class yes { char m;};
class no { yes m[2];};
struct BaseMixin
{
void foo(){}
};
struct Base : public Type, public BaseMixin {};
template <typename T, T t> class Helper{};
template <typename U>
static no deduce(U*, Helper<void (BaseMixin::*)(), &U::foo>* = 0);
static yes deduce(...);
public:
static const bool result = sizeof(yes) == sizeof(deduce((Base*)(0)));
};
struct A {
void foo();
};
struct B : A {};
struct C {};
int main()
{
using namespace std;
cout << boolalpha << has_foo<A>::result << endl;
cout << boolalpha << has_foo<B>::result << endl;
cout << boolalpha << has_foo<C>::result;
}
结果:
true
true
false
关于c++ - SFINAE 检查继承的成员函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/5747205/