C ++返回对临时的引用 [英] C++ Returning reference to temporary

问题描述

可能重复：

警告：返回引用临时

class Object : public std::map <ExString, AnotherObject> const {
public:
const AnotherObject& Find (const ExString& string ) const {
  Object::const_iterator it = find (string);
  if (it == this->end()) { return AnotherObject() };
  return ( it->second );
}
}

我的类实现std :: map。

My class implements std::map.

我是C ++的新手，所以我猜测它只是一个语法错误。任何帮助？

I am new to C++ so I'm guessing its just a syntax error. Any help?

推荐答案

如果您的函数如下所示：

If your function looks like this:

AnotherObject& getAnotherObject()
{

    . . .

    Object::const_iterator it = find ("lang");
    if (it == this->end()) { return AnotherObject() };

    . . .

}

问题是你返回的AnotherObject

the problem is that the AnotherObject() you've returned will be destroyed as soon as the function exits, and so the caller to your function will have a reference to a bogus object.

如果你的函数返回的值是函数返回的值，那么函数返回值为：

If your function returned by value however:

AnotherObject getAnotherObject()

b $ b

那么在原件被销毁之前将会进行复制，你就OK了。

then a copy will be made before the original is destroyed and you'll be OK.

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