是返回一个C ++引用变量的做法吗？ [英] Is the practice of returning a C++ reference variable, evil?

问题描述

这是有点主观的，我想;我不知道这个意见是否一致（我已经看到很多代码片段返回了引用）。

This is a little subjective I think; I'm not sure if the opinion will be unanimous (I've seen a lot of code snippets where references are returned).

根据这个问题我刚才问，关于初始化引用，返回一个引用可以是邪恶因为，[我知道]，它使它更容易错过删除它，这可能导致内存泄漏。

According to a comment toward this question I just asked, regarding initializing references, returning a reference can be evil because, [as I understand] it makes it easier to miss deleting it, which can lead to memory leaks.

这让我担心，因为我遵循的例子我想象的东西），并在一个公平的几个地方做这个...我误解了吗？是邪恶的吗？如果是这样，是多么邪恶？

This worries me, as I have followed examples (unless I'm imagining things) and done this in a fair few places... Have I misunderstood? Is it evil? If so, just how evil?

我觉得因为我混合的指针和引用，加上我是新的C ++的事实，什么时候使用什么时候，我的应用程序必须是内存泄露地狱...

I feel that because of my mixed bag of pointers and references, combined with the fact that I'm new to C++, and total confusion over what to use when, my applications must be memory leak hell...

此外，我理解使用智能/共享指针通常被接受为避免内存泄漏。

Also, I understand that using smart/shared pointers is generally accepted as the best way to avoid memory leaks.

推荐答案

一般来说，返回一个引用是完全正常的，并且一直发生。

In general, returning a reference is perfectly normal and happens all the time.

如果您的意思是：

int& getInt() {
    int i;
    return i;  // DON'T DO THIS.
}

这是各种各样的邪恶。堆栈分配的 i 将会消失，你是指什么。这也是邪恶的：

That is all sorts of evil. The stack-allocated i will go away and you are referring to nothing. This is also evil:

int& getInt() {
    int* i = new int;
    return *i;  // DON'T DO THIS.
}

因为现在客户端最终会做奇怪的事情：

Because now the client has to eventually do the strange:

int& myInt = getInt(); // note the &, we cannot lose this reference!
delete &myInt;         // must delete...totally weird and  evil

int oops = getInt(); 
delete &oops; // undefined behavior, we're wrongly deleting a copy, not the original

请注意，

如果你想分配超出函数范围的东西，使用一个智能指针一般，一个容器）：

If you want to allocate something that lives beyond the scope of the function, use a smart pointer (or in general, a container):

std::unique_ptr<int> getInt() {
    return std::make_unique<int>(0);
}

现在客户端存储一个智能指针：

And now the client stores a smart pointer:

std::unique_ptr<int> x = getInt();

参考文献也可以访问你知道生命在更高级别，例如：

References are also okay for accessing things where you know the lifetime is being kept open on a higher-level, e.g.:

struct immutableint {
    immutableint(int i) : i_(i) {}

    const int& get() const { return i_; }
private:
    int i_;
};

这里我们知道可以返回一个引用 i _ 因为任何调用我们管理类实例的生命周期，所以 i _ 将至少存活这么长。

Here we know it's okay to return a reference to i_ because whatever is calling us manages the lifetime of the class instance, so i_ will live at least that long.

当然，没有什么错误只是：

And of course, there's nothing wrong with just:

int getInt() {
   return 0;
}

如果生命周期应该留给调用者，计算该值。

If the lifetime should be left up to the caller, and you're just computing the value.

摘要：如果对象的生命周期在调用后不会结束，则可以返回引用。

Summary: it's okay to return a reference if the lifetime of the object won't end after the call.

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