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c++stringoperator-overloadingcoutostream

Overload << operator to change " " to "\n"


I am trying to overload

<<

operator. For instance

cout << a << " " << b << " "; // I am not allowed to change this line

is given I have to print it in format

<literal_valueof_a><"\n>
<literal_valueof_b><"\n">
<"\n">

I tried to overload << operator giving string as argument but it is not working. So I guess literal

" "

is not a string. If it is not then what is it. And how to overload it? Kindly help;

Full code

//Begin Program
// Begin -> Non - Editable     
    #include <iostream>
    #include <string>
    using namespace std;

// End -> Non -Editable
//---------------------------------------------------------------------
// Begin -> Editable       (I have written )
    ostream& operator << (ostream& os, const string& str) {
        string s  = " ";
        if(str  ==  " ") {
            os << '\n';
        }
        else {
            for(int i = 0; i < str.length(); ++i)
                os << str[i];
        }
        return os;
    }

// End -> Editable
//--------------------------------------------------------------------------
// Begin -> No-Editable     
int main() {
        int a, b;
        double s, t;
        string mr, ms;
        cin >> a >> b >> s >> t ;
        cin >> mr >> ms ;
        cout << a << " " << b << " " ;
        cout << s << " " << t << " " ;
        cout << mr << " " << ms ;

        return 0;
    }
// End -> Non-Editable
//End Program

Inputs and outputs Input

 30 20 5.6 2.3 hello world 

Output

30
20
5.6
2.3
hello
world

Solution

  • " " is a string-literal of length one, and thus has type const char[2]. std::string is not related.

    Theoretically, you could thus overload it as:

    auto& operator<<(std::ostream& os, const char (&s)[2]) {
        return os << (*s == ' ' && !s[1] ? +"\n" : +s);
    }
    

    While that trumps all the other overloads, now things get really hairy. The problem is that some_ostream << " " is likely not uncommon, even in templates, and now no longer resolves to calling the standard function. Those templates now have a different definition in the affected translation-units than in non-affected ones, thus violating the one-definition-rule.

    What you should do, is not try to apply a global solution to a very local problem:

    Preferably, modify your code currently streaming the space-character.
    Alternatively, write your own stream-buffer which translates it as you wish, into newline.