In the coordinate class, there is a display () member function and a display () const member function with the following code
Class coordinate{public
:
coordinate (int x,int y);
void Display () const;
void Display ();
Private:
int m_ix;
int m_iy;
};
#include <iostream>
#include "Coordinate.h"
using namespace std;
coordinate::coordinate (int x, int y) {
this->m_ix = x;
This->m_iy = y;
}
void coordinate::D isplay () const{
cout << "Display () const" << Endl;
void coordinate::D isplay () {
cout << "Display ()" << Endl;
}
The display () member function and a display () const member function are mutually overloaded, so if we call the method directly as follows, which is the call?
#include <iostream>
#include "Coordinate.h"
using namespace std;
int main () {
coordinate coor (1, 3);
Coor. Display ();
System ("pause");
return 0;
}
Then run the program to see the results
Does the program call a function that has no members modified with const, not that the display () member function and a display () const member function are overloaded, so how do we get the program to call the const-decorated member function?
In fact, it is very simple, just add the const in the declaration of the line.
If there is only one regular member function in the class, the declaration can invoke the constant member function without the addition of a const.
Class coordinate{public
:
coordinate (int x,int y);
void Display () const;
Private:
int m_ix;
int m_iy;
};
#include <iostream>
#include "Coordinate.h"
using namespace std;
coordinate::coordinate (int x, int y) {
this->m_ix = x;
This->m_iy = y;
}
void coordinate::D isplay () const{
cout << "Display () const" << Endl;
#include <iostream>
#include "Coordinate.h"
using namespace std;
int main () {
coordinate coor (1, 3);
Coor. Display ();
System ("pause");
return 0;
}
The above analysis of the member function and the regular member function of the call is small series to share all the content, hope to give you a reference, but also hope that we support the cloud-dwelling community.