c++ - 如果没有 endl，则重载 ostream 运算符段错误

Question

class foo {
    public:
    friend ostream& operator << (ostream &os, const foo &f);
    foo(int n) : a(n) {}
    private:
    vector <int> a;
};

ostream& operator << (ostream &os, const foo &f) {
    for (int i = 0; i < f.a.size(); ++i)
        os << f.a[i] << " ";
    os << endl; // why is this line a must?
}

int main(void) {
    foo f(2);
    cout << f << endl;
    return 0;
}

上面代码中，如果去掉标记的那一行，会出现segment fault错误，谁能解释一下为什么？

Answer 1

ostream& operator << (ostream &os, const foo &f) {
    for (int i = 0; i < f.a.size(); ++i)
        os << f.a[i] << " ";
    os << endl; // why is this line a must?
}

不是强制性的。段错误是因为您没有返回 os

ostream& operator << (ostream &os, const foo &f) {
    for (int i = 0; i < f.a.size(); ++i)
        os << f.a[i] << " ";
    return os; // Here
}

如果您不返回 ostream，则为未定义行为。 endl 正在刷新您的 os。这就是它看起来有效的原因。

编辑:根据 Bo Persson 的说法，为什么它在这种情况下有效

The os << endl; is another operator call that actually returns os by placing it "where a return value is expected" (likely a register). When the code returns another level to main, the reference to os is still there