c++ - 可以函数返回 0 作为引用

Question

I have this snippet of the code

Account& Company::findAccount(int id){
        for(list<Account>::const_iterator i = listOfAccounts.begin(); i != listOfAccounts.end(); ++i){
            if(i->nID == id){
                return *i;
            }
        }
        return 0;
    }

如果我没有找到合适的帐户，这是返回 0 的正确方法吗？ 因为我收到一个错误:

no match for 'operator!' in '!((Company*)this)->Company::findAccount(id)'

我是这样使用的:

if(!(findAccount(id))){
            throw "hey";
          }

提前致谢

Answer 1

没有空引用这样的东西。标准说:

A reference shall be initialized to refer to a valid object or function. [ Note: in particular, a null reference cannot exist in a well-defined program, because the only way to create such a reference would be to bind it to the “object” obtained by dereferencing a null pointer, which causes undefined behavior.