我想提供两种不同的实现,operator>>具体取决于给定类型是否是特殊类型的子类:
class A {};
class B : public A{};
class C {};
template
std::istream& operator>>(std::istream& is,
std::vector::value>::type>& vec)
{
std::cout << "Special case called" << std::endl;
return is;
}
template
std::istream& operator>>(std::istream& is,
std::vector& vec)
{
std::cout << "General case called" << std::endl;
return is;
}
void main(int argc, char **argv)
{
std::vector a;
std::vector b;
std::vector c;
std::stringstream ss("A string");
ss >> a;
ss >> b;
ss >> c;
}
哪个打印
General case called General case called General case called
将第二个运算符定义更改为
templatestd::istream& operator>>(std::istream& is, std::vector ::value>::type>& vec) { std::cout << "General case called" << std::endl; return is; }
不编译因为
error C2678: binary '>>' : no operator found which takes a left-hand operand of type 'std::stringstream'
所以我可能std::enable_if错了.但是什么是正确的？std::vector这里的模板有问题吗？
我不认为std :: enable_if在这里是最想要的位置,我会把它放在返回类型中以启用SFINAE:
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