Formatting objects into strings

Question

I'm coming back to c++ after using Java for a long time. In Java overriding the toString method on an object allows the object to be automatically translated into a string and concatenated to other strings.

class Test {
    public static void main(String[] args) {
        System.out.println(new Test() + " There"); // prints hello there
    }

    public String toString() {
        return "Hello";
    }
}

Is there anything similar that would allow me to stream an object into cout?

cout << Test() << endl;

Bill · Accepted Answer · 2011-09-17 21:43:57Z

5

The equivalent is to overload operator<<:

#include <ostream>

class Test
{
  int t;
};

std::ostream& operator<<(std::ostream& os, const Test& t)
{
   os << "Hello";
   return os;
}

You would then use it like this:

#include <iostream>

int main()
{
  std::cout << Test() << " There" << std::endl;
}

See the code in action: http://codepad.org/pH1CVYPR

edited Sep 17, 2011 at 21:43

answered Sep 17, 2011 at 21:36

Bill

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10 Comments

Matteo Italia Over a year ago

Overload, not override (override implies that an operator<< for ostream and Test already exists and you are overriding it).

K-ballo Over a year ago

You may want to consider defining the operators for streams taking other kind of characters/traits as well.

Bill Over a year ago

@Matteo: Thanks, I used the wrong one in haste and have fixed it.

K-ballo Over a year ago

template< typename Char, typename Traits > std::basic_ostream< Char, Traits >& operator<<(std::basic_ostream< Char, Traits >& os, const Test& t) will work for all kind of output streams, including std::wcout for instance which is the equivalent of std::cout specified for wchar_t types.

K-ballo Over a year ago

@Bill: Besides the fact that your declaration of the operator is a no-op, the obvious difference between your version and mine is that mine only works for basic_ostream types while yours would work for anything, giving a compile time error if the os << t operation is not well formed for that particular os, or worse silently doing who knows what if the expression is valid and os is not an output stream. Also, note that the full implementation for the output insertion is a bit more complex than just `os << "Hello", as pointed out by @Matteo Italia.

|

Matteo Italia · Accepted Answer · 2011-09-17 21:58:50Z

2

The common idiom is to create an overload of operator<< that takes an output stream as the left-hand operand.

#include <iostream>

struct Point
{
    double x;
    double y;
    Point(double X, double Y)
      : x(X), y(Y)
    {}
};

std::ostream & operator<<(std::ostream & Stream, const Point & Obj)
{
    // Here you can do whatever you want with your stream (Stream)
    // and the object being written into it (Obj)
    // For our point we can just print its coordinates
    Stream<<"{"<<Obj.x<<", "<<Obj.y<<"}";
    return Stream; // return the stream to allow chained writes
}

int main()
{
    Point APoint(10.4, 5.6);
    std::cout<<APoint<<std::endl; // this will print {10.4, 5.6}
    return 0;
}

If you want to support streams with other character types (e.g. wchar_t)/template parameters of the streams you have to write different overloads for the various types of streams you want to support, or, if your code is (more or less) independent from such types, you can simply write a template operator<<.

edited Sep 17, 2011 at 21:58

answered Sep 17, 2011 at 21:35

Matteo Italia

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1 Comment

Matteo Italia Over a year ago

@Kerrek: sure, that's what I meant with the "template operator<<".

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