返回“NULL”对象，如果没有找到搜索结果 [英] Return a "NULL" object if search result not found

问题描述

我是C ++的新手，所以我倾向于设计与许多Java-isms，而我在学习。无论如何，在Java中，如果我有一个'search'方法将从集合返回一个对象 T T> 匹配一个特定的参数，我会返回该对象，如果在集合中找不到该对象，我会返回 null 。然后在我的调用函数中，我只需检查 if（tResult！= null）{...}

I'm pretty new to C++ so I tend to design with a lot of Java-isms while I'm learning. Anyway, in Java, if I had class with a 'search' method that would return an object T from a Collection< T > that matched a specific parameter, I would return that object and if the object was not found in the collection, I would return null. Then in my calling function I would just check if(tResult != null) { ... }

C ++，我发现如果对象不存在，我不能返回 null 值。我只想返回一个类型T的指示符，通知调用函数没有找到任何对象。我不想抛出异常，因为它并不是一个特殊的情况。

In C++, I'm finding out that I can't return a null value if the object doesn't exist. I just want to return an 'indicator' of type T that notifies the calling function that no object has been found. I don't want to throw an exception because it's not really an exceptional circumstance.

这是我的代码现在的样子：

This is what my code looks like right now:

class Node {
    Attr& getAttribute(const string& attribute_name) const {
       //search collection
       //if found at i
            return attributes[i];
       //if not found
            return NULL; // what should this be?
    }

private:
    vector<Attr> attributes;
}

如何更改它以便我可以给这种标记？ p>

How can I change it so I can give that kind of marker?

推荐答案

在C ++中，引用不能为null。如果你想要返回null如果没有找到，你需要返回一个指针，而不是一个引用：

In C++, references can't be null. If you want to optionally return null if nothing is found, you need to return a pointer, not a reference:

Attr *getAttribute(const string& attribute_name) const {
   //search collection
   //if found at i
        return &attributes[i];
   //if not found
        return nullptr;
}

否则，如果你坚持通过引用返回，那么你应该抛出异常如果没有找到属性。

Otherwise, if you insist on returning by reference, then you should throw an exception if the attribute isn't found.

（顺便说一下，我有点担心你的方法 const const Attr * 。如果你也可以修改这个属性，你可以重载一个非 - const 方法返回一个非 - const 属性以及。）

(By the way, I'm a little worried about your method being const and returning a non-const attribute. For philosophical reasons, I'd suggest returning const Attr *. If you also may want to modify this attribute, you can overload with a non-const method returning a non-const attribute as well.)

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