[英]operator<< overload in C++ not working as expected
我已经定义了一个类模板MyVector
并重载了operator<<
来打印它所持有的向量。
//MyVector definition
template<int n, typename T>
class MyVector{
private:
vector<T> vetor;
public:
//Several Methods
MyVector operator+(MyVector& v){
//Adds 2 vectors
}
template<typename U, int m>
friend ostream& operator << (ostream& o, MyVector<m,U>& v);
};
template<typename T, int n>
ostream& operator << (ostream& o, MyVector<n,T> &v){
o << "[";
for(auto itr = v.vetor.begin(); itr != v.vetor.end()-1; ++itr){
o << *itr <<", ";
}
o << v.vetor.back() << "]";
return o;
}
对于简单的用例,运算符似乎工作正常,但是当添加 2 个MyVector
实例时, operator<<
抛出:
invalid operands to binary expression
int main(){
MyVector<2,int> v1;
MyVector<2,int> v2;
MyVector<2,int> v3;
v1.add(1,2);
v2.add(3,4);
v3 = v1+v2;
cout << v3 << endl; // --> This prints "[7,11]" to the console
cout << v1+v2 << endl; // --> This throws the exception
}
我在这里缺少什么?
MyVector
上的operator+
返回一个临时值:
MyVector operator+(MyVector& v);
// ^^^^^^^^ temporary
它不能绑定到operator<<
期望的非常量引用:
ostream& operator << (ostream& o, MyVector<m,U>& v);
// ^^^^^^^^^^^^^^ non-const reference
您可以通过接受 const 引用来解决此问题:
ostream& operator << (ostream& o, MyVector<m,U> const & v);
// ^^^^^
同样,您的operator+
应该通过 const 引用来获取它的参数,并且也应该是 const 限定的:
MyVector operator+(MyVector const & v) const;
// ^^^^^ ^^^^^
允许这样的表达式工作:
MyVector<2, int> const v = // ...
std::cout << v + v + v;
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