I get the compiler error
no match for 'operator<<' in 'std::cout << VertexPriority(2, 4u)'
In the main class referred to this operator overloading, but I can't undestand where the error is.
Here there is the operator overloading line, I implemented it inside the class definition.
std::ostream& operator<<(std::ostream& out) const { return out << "Vertex: " << this->vertex << ", Priority: " << this->priority; }
vertex and priority are integer and unsigner integer.
In the main class I'm trying to doing this:
std::cout << VertexPriority(2, 3) << std::endl;
Define it like this:
class VertexPriority {
...
friend std::ostream& operator<< (std::ostream& out, const VertexPriority& vp);
};
std::ostream& operator<< (std::ostream& out, const VertexPriority& vp) {
return out << "Vertex: " << vp.vertex << ", Priority: " << vp.priority;
}
The friend keyword is necessary if VertexPriority::vertex or VertexPriority::priority are not public.
For more help, read this tutorial: http://www.learncpp.com/cpp-tutorial/93-overloading-the-io-operators/
Related
I am writing this code
ostream& operator <<(ostream& out, Box& B){
return B.l +" "+B.b +" "+B.h +endl;
};
The error I get is
Solution.cpp:40:46: error: ‘std::ostream& Box::operator<<(std::ostream&, Box&)’ must have exactly one argument ostream& operator <<(ostream& out, Box& B){ ^
can someone explain what's wrong? I dont understand.
thanks for your help :)
It seems you mean the following
std::ostream & operator <<( std::ostream& out, const Box& B) {
return out << B.l << " " << B.b << " " << B.h;
}
provided that all used in the operator data members are public data members of the class Box. The operator shall be declared and defined outside the class definition.
If one of the used data members is a private data member of the class then the function should be a friend function of the class and shall be declared (and may be defined) in the class definition. For example
class Box
{
//...
friend std::ostream & operator <<( std::ostream& out, const Box& B) {
return out << B.l << " " << B.b << " " << B.h;
}
//...
};
Pay attention to that it is better to remove in the return statement the operand std::endl. In this case 1) the operator will be more flexible because you can output additional information in the same line and 2) this statement
std::cout << box;
will not confuse readers of the code because they will not see the operand std::endl.
Without this operand in the operator definition in the caller of the operator you can write
std::cout << box << std::endl;
and this statement more clear expresses the intention of the programmer.
It should probably be:
std::ostream& operator<<(std::ostream& out, Box const& B){
out << B.l << " " << B.b << " " << B.h << std::endl;
return out;
};
Full code should look like:
#include <iostream>
class Box {
int l;
int b;
int h;
friend std::ostream& operator<<(std::ostream& out, Box const& B);
};
std::ostream& operator<<(std::ostream& out, Box const& B){
out << B.l << " " << B.b << " " << B.h << std::endl;
return out;
}
int main() {
return 0;
}
Demo
To output a type T to std::ostream, you have to declare standalone operator, which returns reference to that stream
std::ostream& operator<<(std::ostream& out, const T& B)
{
out << B.l << " " << B.b << " " << B.h << '\n';
return out;
}
This operation cannot be member operator because member operator use their class as first argument, therefore it might be necessary declare it as a friend of class T.
Avoid using endl in such operators if not extremely necessary, that manipulator calls out.flush(). If you want new line added, use new-line character.
I am trying to overload the ostream operator as a friend in a class to build components of a circuit, however it keeps returning the address.
In a series-circuit class in the file "Circuit_classes.h":
friend ostream& operator<< (ostream& os, series_circuit const& myCircuit);
In the file "Circuit_classes.cpp":
ostream& operator<<(ostream& os, series_circuit const& myCircuit){
os << "Output: " << myCircuit.frequency << endl;
return os;
}
where frequency is defined in the class header file as 60.
In my main program, "AC Circuits.cpp"
vector<shared_ptr<circuit>> circuit_vector;
circuit_vector.push_back(shared_ptr<circuit>(new series_circuit));
cout << circuit_vector[0] << endl;
Output in the command line when the program is run:
0325E180
cout << circuit_vector[0] << endl;
circuit_vector[0] yields a std::shared_ptr which is what is being printed.
You must dereference it to get to the object itself.
cout << *circuit_vector[0] << endl;
To test two std::pair
BOOST_CHECK_EQUAL(std::make_pair(0.0,0.0), std::make_pair(1.0,1.0));
I overloads the operator<< for std::pair
std::ostream& operator<< (std::ostream& os, const std::pair<double,double>& t)
{
return os << "( " << t.first << ", " << t.second << ")";
}
with following error
error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const std::pair<_Ty1,_Ty2>' (or there is no acceptable conversion)
What is wrong?
Open up the std namespace so ADL can find it.
namespace std
{
ostream& operator<< (ostream& os, const pair<double,double>& t)
{
return os << "( " << t.first << ", " << t.second << ")";
}
}
Ok, I figured it out. Name lookup stops when it finds the name it is looking for in the current namespace, which is why it cannot find your operator<< in global scope, since it found operator<< already in namespace boost because boost declares operator<<.
I recommend reading Why can't I instantiate operator<<(ostream&, vector&) with T=vector? which has a good explanation.
I have successfully overloaded the '<<' operator which I believe is referred to as the insertion operator. I have a print function which prints the information of an instance of a card object, how can I call this print function when the operator is used
example:
Card aCard("Spades",8); //generates an 8 of spades card object
aCard.Print(); // prints the suit and value of card
cout << aCard << endl; // will print something successfully but how can I get the same results as if I were to call the print function?
In my implementation file card.cpp I have overloaded the << for use with my card class.
Card.cpp
void Card::Print()
{
std::cout << "Suit: "<< Suit << std::endl;
std::cout << "Value:" << Value << std::endl;
}
std::ostream& operator<<(std::ostream &out, const Card &aCard)
{
Print();//this causes an error in the program
}
Card.h
class Card
{
public:
std::string Suit;
int Value;
Card(){};
Card(std::string S, int V){Suit=S; Value=V};
void Print();
friend std::ostream& operator<<(std::ostream&, const Card&)
};
You do only want one implementation. You could either make a Print function that takes an ostream and performs all of the print logic then call it from Print() and operator<<
void Card::Print()
{
Print(std::cout);
}
std::ostream& operator<<(std::ostream &out, const Card &aCard)
{
Print(out);
}
void Card::Print(std::ostream &out)
{
out << "Suit: "<< Suit << std::endl;
out << "Value:" << Value << std::endl;
return out;
}
Or you could have the operator<< contain the print logic and call operator<< from Print:
void Card::Print()
{
std::cout << *this;
}
std::ostream& operator<<(std::ostream &out, const Card &aCard)
{
out << "Suit: "<< Suit << std::endl;
out << "Value:" << Value << std::endl;
return out;
}
You need aCard.Print() in operator<< not Print()
std::ostream& operator<<(std::ostream &out, const Card &aCard)
{
aCard.Print();
}
You don't say what the error is but basically you're calling a globally defined Print() function or a function that doesn't exist with your code as it stands.
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);
}