为什么编译器报告“operator<< and operator>> recursive on all control path will cause stack overflow”？

Question

#include <iostream>

class fraction {
    int n, d;
public:
    fraction(){}
    fraction(int n, int d) : n(n), d(d) {}
    int getter() { return n, d; }

    friend std::istream& operator>>(std::istream& stream, const fraction& a) {
        stream >> a;
        return stream;
    }

    friend std::ostream& operator<<(std::ostream& stream, const fraction& a) {
        stream << a;
        return stream;
    }

    friend fraction operator*(const fraction& a, const fraction& b) {
        int pN = a.n * b.n;
        int pD = b.n * b.d;
        return fraction(pN, pD);
    }   
};

int main()
{
    fraction f1;
    std::cout << "Enter fraction 1: ";
    std::cin >> f1;

    fraction f2;
    std::cout << "Enter fraction 2: ";
    std::cin >> f2;

    std::cout << f1 << " * " << f2 << " is " << f1 * f2 << '\n'; // note: The result of f1 * f2 is an r-value

    return 0;
}

Compiling error says:

operator<< and operator>> recursive on all paths, function will cause a stack overflow

I don't know what this means. What does it mean by recursive on all paths and which function is gonna cause a stack overflow?

Answer 1

When you run:

stream >> a;

You are calling the same function you are running, ie friend std::istream& operator>>(std::istream& stream, const fraction& a) .

So you will be calling yourself ( recursion ) again, and again, and again... without an end. This, in turn, means that the memory assigned to the stack will be exhausted at some point (because each frame takes some space) and it will cause a stack overflow .

Instead, you have to do something with the fraction argument a , most likely refer to an and ad .