CODEWITHSUNDEEP
c++
LoremIpsum
LoremIpsum

Reputation: 1862

operator overloading("<<"):why the “cout” did not work properly?

I am learning operator overloading recently. I don't know why cout is outputting nothing after I add const keyword for ostream in argument list. Does this have something to do with the addition of that keyword? the code is as follows:

Program1: 

#include<iostream>
#include<string>

using namespace std;

class Foo
{
private:
    string bar;
public:
    Foo(string str);
    friend  const ostream& operator<<(const ostream& o, const Foo& f);
};

Foo::Foo(string str)
{
    this->bar = str;
}
const ostream&  operator<<(const ostream& o, const Foo& f)
{
    o << f.bar;
    return o;
}
int main()
{
    Foo f("HelloWorld");
    cout << f;
}

output:nothing

Program2 :

#include<iostream>
#include<string>


using namespace std;

class Foo
{
private:
    string bar;
public:
    Foo(string str);
    friend  ostream& operator<<(ostream& o, const Foo& f);


};

Foo::Foo(string str)
{
    this->bar = str;
}
ostream&  operator<<(ostream& o, const Foo& f)
{
    o << f.bar;
    return o;
}
int main()
{
    Foo f("HelloWorld");
    cout << f;
}

output:HelloWorld

Upvotes: 0

Views: 233

Answers (1)

VP.
VP.

Reputation: 16725

The problem is caused by const. Declare your friend function as non-const:

friend ostream& operator<<(ostream& o, const Foo& f);

And implementation:

ostream& operator<<(ostream& o, const Foo& f)
{
    o << f.bar;
    return o;
}

I can't understand why does this code compile because operator<< should always change the object. Your code is an undefined behaviour and may cause a segmentation fault.

Upvotes: 3

Related Questions