CODEWITHSUNDEEP
c++oopcout
mortusdetus
mortusdetus

Reputation: 97

OOP C++ first compiler error

Please help this newbie, here's my code:

#include <iostream>
using namespace std;

class Complex {
private:
    float r, i;

public:
    Complex(float rr, float ii) : r(rr), i (ii) {}
    float GiveRe () { return r; }
    float GiveIm () { return i; }
    void Setit (float rr, float ii) {
        r = rr;
        i = ii;
    }
};

Complex a(10, 20);

Complex sumit (Complex &ref) {
     static Complex sum (0, 0);
    sum.Setit(sum.GiveRe() + ref.GiveRe(), sum.GiveIm() + ref.GiveIm());
    return sum;
}

int main () {
    Complex sumvalue = sumit (a);
    cout << sumvalue << endl;
    return 0;
}

error: no match for 'operator<<' in 'std::cout << sumvalue'.

The program should output the sum of a complex number.

Upvotes: 2

Views: 205

Answers (5)

Rudolf M&#252;hlbauer
Rudolf M&#252;hlbauer

Reputation: 2531

A complete minimal example:

#include <iostream>

using namespace std;

class C {
  public: 
    int r, l;
    // if the operator needs access to private fields:
    friend ostream& operator<< (ostream&, const C&);
};

ostream& operator << (ostream& stream, const C& c) {
    stream << c.r << "--" << c.l;
    return stream;
}

int main() {
    C c;
    c.l = 1;
    c.r = 2;
    cout << c << endl;
}

C++ allows you to define operators. The STL uses the << operator for output, and the whole istream/ostream class hierarchy uses this operator to input/output.

Operators are implemented as functions, but always follow a very specific syntax. As in the example, ostream& operator << (ostream&, const MYTYPEHERE&) is the way to define ostream << operators.

When C++ encounters a statement, it has to deduce the types of all operands, and find (quite magically, indeed) a solution to the question: given my operands and operators, can i find a typing such that the statement gets valid?

These ofstream operators are defined for all basic types somewhere in <iostream>, so if you write cout << 10, the compiler finds an operator ostream& operator<< (ostream&, int).

If you want to be able to use userdefined types in this game, you have to define the operators. otherwise, a statement cout << sometype will not be valid. This is also the reason for the harsh compiler errors sometimes found in C++: "Well, i have some operators << defined, but none is compatible with your type!".

Because sometimes it is not favourable to implement operators for your types (if you only output them once, e.g.), i suggested to write: 

cout << sum.re << "--" << sum.im << endl; // or similar

This way, you write less code, and you are flexible in the output format. Who knows if you want you complex number formatted differently next time? But this is another discussion.

Why complicating that much? Because C++ can be awfully complicated. It it very powerfull, but crammed with special syntax and exceptions to those. In the end, the difference to C lies exactly here: C++ does a much better job with type inference (needed for templates), often resulting in WTF?

On how to implement it in your code, i think the other answers provide nice solutions!

Upvotes: 1

jt234
jt234

Reputation: 672

cout can't tell what you want to output, you need to specify the operator<< in the class or make it possible to implicitly convert your class to a compatible type.

http://www.cplusplus.com/reference/iostream/ostream/operator%3C%3C/

Rudolf Mühlbauer's code as implemented in your class:

Add this somewhere within the class header:

friend ostream& operator<<(ostream& out, const Complex& compl);

and this below the header:

ostream& operator<<(ostream& out, const Complex& compl)
{
    return out << compl.r << "/" << compl.i;
}

Implementation should be changed to suit your exact needs.

Upvotes: 2

Aniket Inge
Aniket Inge

Reputation: 25705

You have to overload the "<<" operator for Complex type.

#include <iostream>
using namespace std;

class Complex {
private:
    float r, i;

public:
    Complex(float rr, float ii) : r(rr), i (ii) {}
    float GiveRe () { return r; }
    float GiveIm () { return i; }
    void Setit (float rr, float ii) {
        r = rr;
        i = ii;
    }

};

ostream& operator<<(ostream& os, Complex& c)
    {
        float i;
        os<<c.GiveRe();
        if(c.GiveIm() < 0){  
            os<<"-j"<<c.GiveIm()*(-1);
        }else{
            os<<"+j"<<c.GiveIm();
        }
        return os;
    }

Complex a(10, 20);

Complex sumit (Complex &ref) {
     static Complex sum (0, 0);
    sum.Setit(sum.GiveRe() + ref.GiveRe(), sum.GiveIm() + ref.GiveIm());
    return sum;
}

int main () {
    Complex sumvalue = sumit (a);
    cout << sumvalue << endl;
    return 0;
}

Upvotes: 1

EddieV223
EddieV223

Reputation: 5303

Complex doesn't have an operator <<

ostream& Complex::operator << ( ostream& os )
    {
    // use os << field/method here to out put
    return os;
    }

Also if complex can be displayed to console in different ways, then you should think of using methods to display instead of cout << 

void Complex::DisplayToConsole()
    {
    std::cout << r << " " << i << '\n';
    }

Upvotes: 1

David G
David G

Reputation: 96835

sumit(a) returns an object of type Complex, which cout was not defined to handle.

Upvotes: 0

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