我在vrkvector.h文件中编写了如下所示的简单类。
namespace vrk {
class VRKVector {
private:
std::vector<int> m_coordinates;
int m_dimension;
public:
/** VRKVector constructor.
* @param coordinates wich represents a vector coordinates.
* @return none.
*/
VRKVector(std::vector<int> coordinates);
/** operator << : output stream operator
* @param out which is output stream we want to write
* @param vector wich represents a vector coordinates.
* @return none.
*/
friend std::ostream& operator << (std::ostream& out, vrk::VRKVector& vrkvector);
};
}
VRKVector.cpp
// ============================================================================
// Local includes, e.g. files in the same folder, typically corresponding declarations.
#include "VRKVector.h"
// ============================================================================
// System includes, e.g. STL.
#include <iostream>
#include <fstream>
#include <sstream>
#include <iterator>
#include <algorithm>
// namespace declarations;
using namespace std;
using namespace vrk;
VRKVector::VRKVector(std::vector<int> coordinates) {
m_coordinates = coordinates;
m_dimension = coordinates.size();
}
std::ostream& operator<< (std::ostream& out, vrk::VRKVector& vrkvector) {
out << vrkvector.m_dimension;
return out; // return std::ostream so we can chain calls to operator<<
}
上面的代码我得到了如下所示的错误
1>C:\VRKVector.cpp(42,18): error C2248: 'vrk::VRKVector::m_dimension': cannot access private member declared in class 'vrk::VRKVector'
1>C:\\VRKVector.h(40): message : see declaration of 'vrk::VRKVector::m_dimension'
我们可以评估私人成员，因为我们有朋友功能。 为什么我看到了错误。
谢谢
您需要在命名空间vrk中定义运算符<<<，friend声明将其声明为命名空间vrk的成员。
在类或类模板X内的友元声明中首次声明的名称将成为X的最内部封闭命名空间的成员，。。。
std::ostream& vrk::operator<< (std::ostream& out, vrk::VRKVector& vrkvector) {
// ^^^^^
out << vrkvector.m_dimension;
return out; // return std::ostream so we can chain calls to operator<<
}
std::ostream& operator<< (std::ostream& out, vrk::VRKVector& vrkvector){}
不是的定义
friend std::ostream &operator<<(std::ostream &out, vrk::VRKVector &vrkvector);
由于它位于不同的命名空间中，因此它无法访问类私有成员，这是试图访问给定类私有成员的全局函数所期望的。
您可以通过以下方法修复:
friend std::ostream &operator<<(std::ostream &out, vrk::VRKVector &vrkvector){
out << vrkvector.m_dimension;
return out; // return std::ostream so we can chain calls to operator<<
}
>
在命名空间内定义它。
使用命名空间范围:
std::ostream& vrk::operator<< (std::ostream& out, vrk::VRKVector& vrkvector) {
/*...*/
}
