why can't we call a function on a const object?

Question

the problem is in main where I want to call set() function on display() function which return a reference to const.

#include "stdafx.h"
#include "iostream"
#include "string"

using std::cin;
using std::cout;
using std::endl;
using std::string;
using std::ostream;

class screen
{
public:
typedef string::size_type pos;
screen() = default;
screen(pos ht, pos wd, char c) :height(ht), width(ht),contents(ht*wd, c){}
screen(pos ht, pos wd) :height(ht), width(wd), contents(ht*wd, ' '){}
char get()const{ return contents[cursor]; }
inline char get(pos r, pos c)const;
screen &move(pos r, pos c);
screen &set(char);
screen &set(pos r, pos c,char);
const screen& display(ostream &os)const;
    pos cursor;
    pos height, width;
    string contents;
};
const screen& screen::display(ostream &os)const
{
     os<< contents;
     return *this;
}

char screen::get(pos r, pos c)const
{
    pos row = r*width;
    return contents[row + c];
}
screen& screen::move(pos r, pos c)
{
    pos row = r*width;
    cursor = row + c;
    return *this;
}
screen& screen::set(char c)
{
    contents[cursor] = c;
    return *this;
}
screen& screen::set(pos r, pos c, char ch)
{
    pos row = r*width;
    contents[row + c] = ch;
    return *this;
}

int main()
{
    screen myscreen(50, 50,'0');
    myscreen.display(cout).set('#');//the problem is here!


    return 0;
}

I just want to know why set('#') cannot be called on display(cout) which returns a reference to const. It is possible to set('#').display(cout) but not display(cout).set('#') How constness of an abject can have such a effect on a call?

Answer 1

If you call a member function, there is an implicit this pointer passing as a parameter to the function. In your case, the this pointer is const, and it can't be modified inside your set function. So it failed.