如果没有endl，重载ostream运算符分段错误 [英] Overloaded ostream operator segmentation fault if no endl

问题描述

  class foo {
 public：
 friend ostream&运算符<< （ostream& os，const foo& f）; 
 foo（int n）：a（n）{} 
 private：
 vector< int>一个; 
}; 
 
 ostream&运算符<< （ostream& os，const foo& f）{
 for（int i = 0; i  os< f.a [i]<< ; 
 os<< endl //为什么这行是必须的？ 
} 
 
 int main（void）{
 foo f（2）; 
 cout<< f<< endl 
 return 0; 
} 
  

在上面的代码中，如果标记行被删除，段错误，可以解释为什么吗？

解决方案

 运算符<< （ostream& os，const foo& f）{
 for（int i = 0; i  os< f.a [i]<< ; 
 os<< endl //为什么这行是必须的？ 
} 
  

不是manadatory。 segfault是因为你没有返回 os

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

如果不返回ostream， endl 在此处刷新您的 os 。

b b b b b b b blockquote>

os< endl是另一个操作符调用，它实际上通过
返回os，将它置于预期返回值（可能是寄存器）。
当代码向main返回另一个级别时，对os的引用是
仍然存在

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;
}

In the above code, if the marked line is removed, there will be a segment fault error, can someone explain why?

解决方案

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?
}

is not manadatory. The segfault is caused because you are not returning os

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

it is undefined behavior if you don't return the ostream. The endl is flushing your os here. That's why it seems like it is working.

EDIT: Why it is working in this case according to Bo Persson

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

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