朋友声明*

Question

连接到这个问题的代码编译失败（克++ 4.7）与以下错误：

test.cpp:22:31: error: declaration of ‘operator*’ as non-function 
test.cpp:22:31: error: expected ‘;’ at end of member declaration 
test.cpp:22:32: error: expected unqualified-id before ‘<’ token 
test.cpp: In function ‘int main(int, char**)’: 
test.cpp:39:11: warning: unused variable ‘d’ [-Wunused-variable] 
test.cpp: In instantiation of ‘Vector3<T> operator*(T, const Vector3<T>&) [with T = float]’: 
test.cpp:37:16: required from here 
test.cpp:25:10: error: ‘float Vector3<float>::v [3]’ is private 
test.cpp:30:57: error: within this context 
test.cpp:25:10: error: ‘float Vector3<float>::v [3]’ is private 
test.cpp:30:57: error: within this context 
test.cpp:25:10: error: ‘float Vector3<float>::v [3]’ is private 
test.cpp:30:57: error: within this context 

然而，如果朋友声明之前所移动的operator*成员声明/定义（简单地交换在示例代码的注释），那么它编译罚款。我不明白为什么。有没有什么办法可以使用位于operator*成员声明之后的朋友声明来编译此代码？

template<typename T> 
class Vector3; 

template<typename T> 
Vector3<T> operator*(T f, const Vector3<T>& v); 

template<typename T> 
class Vector3 { 
public: 
    Vector3(T x, T y, T z) { 
     v[0] = x; 
     v[1] = y; 
     v[2] = z; 
    } 

    //friend Vector3<T> operator*<>(T f, const Vector3<T>& v); // WORKS 

    T operator*(const Vector3<T>& other) const { 
     return (v[0] * other.v[0] + v[1] * other.v[1] + v[2] * other.v[2]); 
    } 

    friend Vector3<T> operator*<>(T f, const Vector3<T>& v); // FAILS 

private: 
    T v[3]; 
}; 

template<typename T> 
Vector3<T> operator*(T f, const Vector3<T>& v) { 
    return Vector3<T>(f * v.v[0], f * v.v[1], f * v.v[2]); 
} 

int main(int argc, char *argv[]) { 
    Vector3<float> v(0, 0, 0); 
    Vector3<float> w(0, 0, 0); 

    w = 2.0f * v; 

    float d = v * w; 

    return 0; 
} 

Answer 1

的问题是，你没有指定你的意思是从全局命名空间的operator *，要做到这一点您的朋友声明更改为：

friend Vector3<T> (::operator*<>)(T f, const Vector3<T>& v); 

当你面前的operator *成员宣布它有没有歧义，所以它被正确地拾起。