I am really confused while overloading << operator as friend function.
This line works fine
cout << endl;
But this line gives compilation issue. Why ??
operator <<(cout, endl);
Below is the sample code
class Logger
{
int _id;
string _name;
public:
Logger(int id, string name):_id(id),_name(name){
cout<<"Constructor"<<endl;
}
~Logger(){
cout<<"destructor"<<endl;
}
friend ostream& operator <<( ostream& out,Logger& log);
};
ostream& operator << (ostream& out,Logger& log)
{
out<<log._id<<" "<<log._name;
return out;
}
And What is the need of return statment ? Without return also the below statment works fine.
cout<< log << endl << endl << log2 << endl << log3 << endl;
The operator<< that takes a stream manipulator is a member function of basic_ostream. You can't call it as if it were a free function; you have to call it as a member function:
std::cout.operator<<(std::endl)
On the other hand, the stream inserter that takes a std::string is a free function, and you can call it with the usual function call:
std::string text = "Hello, world";
operator<<(std::cout, text);
but not as a member function.
std::endl is usually the wrong thing to use; '\n' ends a line, without the extra stuff that std::endl does.
Related
Error ocurred with the following try to operator overloading:
#include<iostream>
#include<string>
#include<ostream>
using namespace std;
class Dollar
{
private:
float currency, mktrate, offrate;
public:
Dollar(float);
float getDollar() const;
float getMarketSoums() const;
float getofficialSoums() const;
void getRates();
// In the following function I was trying to overload "<<" in order to print all the data members:
friend void operator<<(Dollar &dol, ostream &out)
{
out << dol.getDollar() << endl;
out << dol.getMarketSoums() << endl;
out << dol.getofficialSoums() << endl;
}
};
Dollar::Dollar(float d)
{
currency = d;
}
float Dollar::getDollar() const
{
return currency;
}
float Dollar::getMarketSoums() const
{
return mktrate;
}
float Dollar::getofficialSoums() const
{
return offrate;
}
void Dollar::getRates()
{
cin >> mktrate;
cin >> offrate;
}
int main()
{
Dollar dollar(100);
dollar.getRates();
// In this line I am getting the error. Could you please help to modify it correctly?
cout << dollar;
system("pause");
return 0;
}
You have to pass std::ostream object as the first parameter to the insertion operator << not as the second one as long as you are calling it that way:
friend void operator << (ostream &out, Dollar &dol);
You should make the object passed in to the insertion operator constant reference as long as this function is only prints and not intending to modify the object's members:
friend void operator << (ostream &out, const Dollar& dol);
So pass by reference to avoid multiple copies and const to avoid unintentional modification.
If you want to invoke to get it work the way you wanted you can do this:
friend void operator<<(const Dollar &dol, ostream &out){
out << dol.getDollar() << endl;
out << dol.getMarketSoums() << endl;
out << dol.getofficialSoums() << endl;
}
And in main for example:
operator << (dollar, cout); // this is ok
dollar << cout; // or this. also ok.
As you can see I reversed the order of calling the insertion operator to match the signature above. But I don't recommend this, it is just to understand more how it should work.
I'm having troubles understanding the reason why the compiler accuses error, when the return type of a << operator overload is std::string. Could you please help me understand?
Bellow is an reproducible example, which gives a gigantic error.
class XY
{
int X__;
int Y__;
public:
XY(int x, int y):X__(x), Y__(y){}
~XY(){}
std::string operator<<(const XY_cartesiano& c)
{
std::stringstream ss;
ss << "{ " << X__ << ", " << Y__ << " }";
return ss.str();
}
int x() const{return X__;}
int y() const{return Y__;}
};
void main()
{
XY a(1,2);
std::cout << a;
}
Let's take something like this as an example:
cout << "My number is " << 137 << " and I like it a lot." << endl;
This gets parsed as
((((cout << "My number is ") << 137) << " and I like it a lot.") << endl);
In particular, notice that the expression cout << "My number is " has to evaluate to something so that when we then try inserting 137 with << 137 the meaning is "take 137 and send it to cout."
Imagine if cout << "My number is " were to return a string. In that case, the << 137 bit would try to use the << operator between a string on the left-hand side and an int on the right-hand side, which isn't well-defined in C++.
The convention is to have the stream insertion operator operator << return a reference to whatever the left-hand side stream is so that these operations chain well. That way, the thing on the left-hand side of << 137 ends up being cout itself, so the above code ends up essentially being a series of chained calls to insert things into cout. The signature of these functions therefore usually look like this:
ostream& operator<< (ostream& out, const ObjectType& myObject) {
// ... do something to insert myObject into out ... //
return out;
}
Now, everything chains properly. Notice that this function is a free function, not a member function, and that the left-hand side is of type ostream and the right-hand side has the type of your class in it. This is the conventional way to do this, since if you try overloading operator << as a member function, the left-hand side will be an operand of your class type, which is backwards from how stream insertion is supposed to work. If you need to specifically access private fields of your class in the course of implementing this function, make it a friend:
class XY {
public:
...
friend ostream& operator<< (ostream& out, const XY& myXY);
};
ostream& operator<< (ostream& out, const XY &myXY) {
...
return out;
}
Correct way to overload << operator in your case is
ostream& operator<<(ostream& os, const XY& c)
{
os << c.X__ <<" "<< c.Y__ ;
return os;
}
You have overloaded operator<< in a way that's incompatible with the conventions you must follow when you intend to use the operator with a std::ostream object like std::cout.
In fact, your operator<<'s signature has nothing to do with streams at all! It is just a member function of XY which takes another XY (which it then does not use), returns a string and has an unsual name. Here's how you would theoretically call it:
XY a(1,2);
XY b(1,2);
std::string x = (a << b);
The correct way to overload operator<< for use with streams is to make the operator a non-member function, add a stream reference parameter and return a stream reference to the stream argument. You also do not need a string stream; you write directly to the stream you get:
#include <iostream>
class XY
{
int x;
int y;
public:
XY(int x, int y) : x(x), y(y) {}
int X() const { return x; }
int Y() const { return y; }
};
std::ostream& operator<<(std::ostream& os, XY const& c)
{
os << "{ " << c.X() << ", " << c.Y() << " }";
return os;
}
int main()
{
XY a(1,2);
std::cout << a;
}
Everywhere I look I see everyone overloading the << operator in a few lines, it seems very simple, but for some reason, overloading the << operator in my code does nothing.
In my .h I have:
friend std::ostream& operator<<(std::ostream& os, const Test& test);
And in my .cpp I have:
std::ostream& operator<<(std::ostream& out,const DeckOfCards& deck) {
out << "oi"; //just testing with a normal string before i try methods
return out;
}
And finally in the main function I have:
Test* test = new Test();
std::cout << "output is: " << test << std::endl;
Could someone please tell me what I'm doing wrong?
Thanks in advance.
How about trying this:
std::cout << "output is: " << *test << std::endl;
In your code, you are couting the pointer, not the object.
I learnt how to do operator overloading of Stream Insertion Operator. But one doubt remains.
#include<iostream>
class INT
{
int i;
friend std::ostream& operator<<(std::ostream&,INT&);
public:
INT():i(100){}
};
std::ostream& operator<<(std::ostream& obj,INT & data)
{
obj<<data.i;
return obj;
}
int main()
{
INT obj;
std::cout<<obj;
}
What significance return obj; has?
Do that return have any use further?
Are We forced to do that return because of the syntax of the operator<< without any usefulness?
Remember how you can write code like this:
cout << "The data is: " << somedata << endl;
This is actually the same as:
((cout << "The data is: ") << somedata) << endl;
For this to work, the << operator has to return the stream.
I want to overload the << operator for my class Complex.
The prototype is the following:
void Complex::operator << (ostream &out)
{
out << "The number is: (" << re <<", " << im <<")."<<endl;
}
It works, but I have to call it like this: object << cout for the standart output.
What can I do to make it work backwards, like cout << object?
I know that the 'this' pointer is by default the first parameter sent to the method so that's why the binary operator can work only obj << ostream. I overloaded it as a global function and there were no problems.
Is there a way to overload the << operator as a method and to call it ostream << obj?
I would just use the usual C++ pattern of free function. You can make it friend to your Complex class if you want to make Complex class's private data members visible to it, but usually a complex number class would expose public getters for real part and imaginary coefficient.
class Complex
{
....
friend std::ostream& operator<<(std::ostream &out, const Complex& c);
private:
double re;
double im;
};
inline std::ostream& operator<<(std::ostream &out, const Complex& c)
{
out << "The number is: (" << c.re << ", " << c.im << ").\n";
return out;
}
You could write a free stand operator<< function, try:
std::ostream& operator<< (std::ostream &out, const Complex& cm)
{
out << "The number is: (" << cm.re <<", " << cm.im <<")." << std::endl;
return out;
}
You can define a global function:
void operator << (ostream& out, const Complex& complex)
{
complex.operator<<(out);
}