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;
}
解决方法
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?
}
不是躁狂的.由于您没有返回操作系统而导致段错误
ostream& operator << (ostream &os,const foo &f) {
for (int i = 0; i < f.a.size(); ++i)
os << f.a[i] << " ";
return os; // Here
}
如果你不返回ostream,它是未定义的行为. endl在这里冲你的操作系统.这就是它似乎正在起作用的原因.
编辑：根据Bo PeRSSon,为什么它在这种情况下工作
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
