why is this code not detecting << operator? - c++

Question

Please explain to me why this is not detecting the << operator.
I tried my best, even tried to overload << on both classes (which is not necessary).
#include<iostream>
using namespace std;
const int MAX = 10;
class Complex;
template<class t>
class stack {
private:
t stk[MAX];
int top;
public:
stack() { top = -1; }
void push(t data) {
if (top == MAX - 1)
cout << "Stack is full.";
else
stk[++top] = data;
}
t pop() {
if (top == -1) {
cout << "Stack is empty.";
return NULL;
}
else {
//return stk[top--];
t data = stk[top];
top--;
return data;
}
}
};
class Complex {
private:
float real, imag;
public:
Complex(float r = 0.0, float i = 0.0) { real = r; imag = i; }
friend ostream& operator << (ostream& s, Complex& c);
};
ostream& operator << (ostream& s, Complex& c) {
s << "(" << c.real << "," << c.imag << ")";
return s;
}
int main() {
stack<int> s1;
s1.push(10);
s1.push(20);
s1.push(30);
s1.push(40);
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
stack<float> s2;
s2.push(3.14f);
s2.push(4.14f);
s2.push(5.14f);
s2.push(6.14f);
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
Complex c1(1.5f, 2.5f), c2(1.5f, 2.5f), c3(1.5f, 2.5f), c4(1.5f, 2.5f);
//cout<<c1;
stack<Complex> s3;
s3.push(c1);
s3.push(c2);
s3.push(c3);
s3.push(c4);
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
return 0;
}

Answer 1

The function signature should be something like:
std::ostream& operator<<(std::ostream& os, const T& obj)
{
// write obj to stream
return os;
}
And the function pop can't return NULL since the type is not the same as t. Fixed code might like:
#include <iostream>
using namespace std;
const int MAX = 10;
class Complex;
template <class t>
class stack {
private:
t stk[MAX];
int top;
public:
stack() { top = -1; }
void push(t data) {
if (top == MAX - 1)
cout << "Stack is full.";
else
stk[++top] = data;
}
t pop() {
if (top == -1) {
cout << "Stack is empty.";
return {}; // May throw, or return std::optional here
} else {
// return stk[top--];
t data = stk[top];
top--;
return data;
}
}
};
class Complex {
private:
float real, imag;
public:
Complex(float r = 0.0, float i = 0.0) {
real = r;
imag = i;
}
friend ostream &operator<<(ostream &s, const Complex &c);
};
ostream &operator<<(ostream &s, const Complex &c) {
s << "(" << c.real << "," << c.imag << ")";
return s;
}
int main() {
stack<int> s1;
s1.push(10);
s1.push(20);
s1.push(30);
s1.push(40);
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
cout << s1.pop() << endl;
stack<float> s2;
s2.push(3.14f);
s2.push(4.14f);
s2.push(5.14f);
s2.push(6.14f);
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
cout << s2.pop() << endl;
Complex c1(1.5f, 2.5f), c2(1.5f, 2.5f), c3(1.5f, 2.5f), c4(1.5f, 2.5f);
// cout<<c1;
stack<Complex> s3;
s3.push(c1);
s3.push(c2);
s3.push(c3);
s3.push(c4);
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
cout << s3.pop() << endl;
return 0;
}
Online demo.
Related question: What are the basic rules and idioms for operator overloading?

Answer 2

It doesn't work because the second parameter should be a const reference
class Complex{
private:
float real,imag;
public:
Complex(float r=0.0,float i=0.0){
real=r;
imag=i;
}
friend ostream& operator<<(ostream& s, const Complex& c);
};
ostream& operator<<(ostream& s, const Complex& c) {
s<<"("<<c.real<<","<<c.imag<<")";
return s;
}

Answer 3

Reason:
The pop function returns an r-value. You can think of an r-value as a nameless object in memory (of course, there is more to it). You cannot put an r-value into an l-value reference. But for a const reference, it does not matter, since you are not going to modify it.
Solution 1:
friend ostream &operator<<(ostream &s, Complex &c); // accepts lvalue - c refers to some external Complex object.
friend ostream &operator<<(ostream &s, Complex &&c); // accepts rvalue - c becomes a normal local variable whose value is the passed rvalue.
Solution 2:
friend ostream &operator<<(ostream &s, const Complex &c); // accepts both