c++ - 链接器错误 : Template relational operator

Question

为什么这段代码会给我一个链接器错误，我该如何解决？

架构 x86_64 的 undefined symbol :“operator==(foo const&, foo const&)”，引用自:_main in main.o ld: symbol(s) not found for architecture x86_64

template<typename T>
class foo {
  //friends get access to the private member t
  friend bool operator==(const foo<T> &lhs, const foo<T> &rhs);
  T t;
};

template<typename T>
bool operator==(const foo<T> &lhs, const foo<T> &rhs) {
  return lhs.t == rhs.t;
}

int main(int,char**) {
  foo<int> f1, f2;
  if (f1 == f2)
    ;
  return 0;
}

Answer 1

这是您的代码的修复:

template<typename T>
class foo; // Forward declaration

template<typename T> // Need to define or declare this before the class
bool operator==(const foo<T> &lhs, const foo<T> &rhs) {
  return lhs.t == rhs.t; 
}

template<typename T>
class foo {
  // Notice the little <> here denoting a specialization
  friend bool operator==<>(const foo<T> &lhs, const foo<T> &rhs);
  T t;
};