C++ 从迭代器返回对象的引用
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return reference of an object from an iterator
提问by hakuna matata
I want to return a reference of an object from a vector, and the object is in an iterator object. How can I do that?
我想从一个向量返回一个对象的引用，并且该对象在一个迭代器对象中。我怎样才能做到这一点？
I tried the following:
我尝试了以下方法：
Customer& CustomerDB::getCustomerById (const string& id) {
vector<Customer>::iterator i;
for (i = customerList.begin(); i != customerList.end() && !(i->getId() == id); ++i);
if (i != customerList.end())
return *i; // is this correct?
else
return 0; // getting error here, cant return 0 as reference they say
}
In the code, customerList is a vector of customers, and the function getId returns the id of the customer.
代码中customerList是一个客户向量，函数getId返回客户的id。
Is the *icorrect? And how can I return 0 or null as a reference?
是*i正确的吗？以及如何返回 0 或 null 作为参考？
回答by reko_t
return *i;is correct, however you can't return 0, or any other such value. Consider throwing an exception if the Customer is not found in the vector.
return *i;是正确的，但是您不能返回 0 或任何其他此类值。如果在向量中找不到 Customer，请考虑抛出异常。
Also be careful when returning references to elements in vector. Inserting new elements in vector can invalidate your reference if vector needs to re-allocate its memory and move the contents.
返回对向量中元素的引用时也要小心。如果 vector 需要重新分配其内存并移动内容，则在 vector 中插入新元素会使您的引用无效。
回答by suszterpatt
There is no such thing as a "null" reference: if your method gets an id that's not in the vector, it will be unable to return any meaningful value. And as @reko_t points out, even a valid reference may become invalid when the vector reallocates its internals.
没有“空”引用这样的东西：如果你的方法得到一个不在向量中的 id，它将无法返回任何有意义的值。正如@reko_t 指出的那样，当向量重新分配其内部结构时，即使是有效的引用也可能变得无效。
You should only use reference return types when you can always return a reference to an existing object that will stay valid for some time. In your case, neither is guaranteed.
仅当您始终可以返回对将保持有效一段时间的现有对象的引用时，才应使用引用返回类型。在您的情况下，两者都不能保证。
