一，判断在类中某个函数(也可以是变量或类型)是否存在
template<typename T>
struct xxxx_detector
{
template<typename P,void (P::*)(void)> struct detector{};
template<typename P> static char func(detector<P,&P::init>*);
template<typename P> static long func(...);
static constexpr bool value = sizeof(func<T>(nullptr)) == sizeof(char);
};
这样的话xxxx_detector<YYY>::value，当类型YYY存在成员函数void YYY::init(void)，这个值就是true。
template<typename T>
struct has_hello
{
template<typename U, void (U::*)()> struct HELPS;
template<typename U> static char Test(HELPS<U, &U::hello>*);
template<typename U> static int Test(...);
const static bool value = sizeof(Test<T>(0)) == sizeof(char);
};
template<typename T>
struct hello_detector
{
template<typename P,void (P::*)(void)> struct detector{};
template<typename P> static char func(detector<P,&P::hello>*);
template<typename P> static long func(...);
static constexpr bool value = sizeof(func<T>(nullptr)) == sizeof(char);
};
struct A
{
void hello(){
cout<<"A is Hello."<<endl;
}
int x;
};
int main()
{
cout<<"A has hello? "<<has_hello<A>::value<<endl;
cout<<"A has hello? "<<hello_detector<A>::value<<endl;
}