Cpp - const Pointers and const Member Functions

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Introduction

If you declare a pointer to a const object, the only functions that you can call with that pointer are const functions.

When you declare an object to be const, you are declaring that the this pointer is a pointer to a const object.

A const this pointer can be used only with const member functions.

Demo

#include <iostream> 
 
class Rectangle //ww  w.ja v  a  2  s.c om
{ 
public: 
    Rectangle(); 
    ~Rectangle(); 
    void SetLength(int length) { itsLength = length; } 
    int GetLength() const { return itsLength; } 
 
    void SetWidth(int width) { itsWidth = width; } 
    int GetWidth() const { return itsWidth; } 
 
private: 
    int itsLength; 
    int itsWidth; 
}; 
 
Rectangle::Rectangle(): 
itsWidth(5), 
itsLength(10) 
{} 
 
Rectangle::~Rectangle() 
{} 
 
int main() 
{ 
    Rectangle* pRect =  new Rectangle; 
    const Rectangle *pConstRect = new Rectangle; 
    Rectangle * const pConstPtr = new Rectangle; 
 
    std::cout << "pRect width: " 
                  << pRect->GetWidth() << " feet\n"; 
    std::cout << "pConstRect width: " 
                  << pConstRect->GetWidth() << " feet\n"; 
    std::cout << "pConstPtr width: " 
                  << pConstPtr->GetWidth() << " feet\n"; 
   
    pRect->SetWidth(10); 
    // pConstRect->SetWidth(10); 
    pConstPtr->SetWidth(10); 
   
    std::cout << "pRect width: " 
                  << pRect->GetWidth() << " feet\n"; 
    std::cout << "pConstRect width: " 
                  << pConstRect->GetWidth() << " feet\n"; 
    std::cout << "pConstPtr width: " 
                  << pConstPtr->GetWidth() << " feet\n"; 
    return 0; 
}

Result

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