此代码似乎实现在C ++中返回null引用 [英] This code appears to achieve the return of a null reference in C++

问题描述

我的C ++知识有点零散。我在工作中重做一些代码。我改变了一个函数来返回一个类型的引用。在内部，我基于传入的标识符查找对象，然后返回对该对象的引用（如果找到）。当然，我遇到了如果我没有找到对象，返回的问题，在环顾四周，许多人声称在C ++中返回一个null引用是不可能的。基于这个建议，我尝试了返回一个成功/失败布尔值，并使对象引用一个out参数的把戏。但是，我遇到了需要初始化我将作为实际参数传递的引用的路障，当然没有办法这样做。我退回到通常的方法只是返回一个指针。

My C++ knowledge is somewhat piecemeal. I was reworking some code at work. I changed a function to return a reference to a type. Inside, I look up an object based on an identifier passed in, then return a reference to the object if found. Of course I ran into the issue of what to return if I don't find the object, and in looking around the web, many people claim that returning a "null reference" in C++ is impossible. Based on this advice, I tried the trick of returning a success/fail boolean, and making the object reference an out parameter. However, I ran into the roadblock of needing to initialize the references I would pass as actual parameters, and of course there is no way to do this. I retreated to the usual approach of just returning a pointer.

我问一个同事。他经常使用以下技巧，这是最近版本的Sun编译器和gcc接受的：

I asked a colleague about it. He uses the following trick quite often, which is accepted by both a recent version of the Sun compiler and by gcc:

MyType& someFunc(int id)
{
  // successful case here:
  // ...

  // fail case:
  return *static_cast<MyType*>(0);
}

// Use:

...
MyType& mt = somefunc(myIdNum);

if (&mt) // test for "null reference"
{
  // whatever
}
...

我一直在维护这个代码库，但我发现我没有足够的时间看关于语言的小细节，我想。我一直在浏览我的参考书，但对此的答案逃避了我。

I have been maintaining this code base for a while, but I find that I don't have as much time to look up the small details about the language as I would like. I've been digging through my reference book but the answer to this one eludes me.

现在，我几年前有一个C ++课程，在那里我们强调在C ++中，一切都是类型，所以我在尝试保持这一点。解构表达式：* static_cast< MyType *>（0）;，在我看来，我们取一个字面值零，将它转换为一个指向MyType的指针（使它成为一个空指针），然后应用在分配给引用类型（返回类型）的上下文中，它应该给我一个对指针指向的同一对象的引用。

Now, I had a C++ course a few years ago, and therein we emphasized that in C++ everything is types, so I try to keep that in mind when thinking things through. Deconstructing the expression: "*static_cast<MyType*>(0);", it indeed seems to me that we take a literal zero, cast it to a pointer to MyType (which makes it a null pointer), and then apply the dereferencing operator in the context of assigning to a reference type (the return type), which should give me a reference to the same object pointed to by the pointer. This sure looks like returning a null reference to me.

任何建议解释为什么这是有效的（或为什么它不应该）。

Any advice in explaining why this works (or why it shouldn't) would be greatly appreciated.

谢谢，
Chuck

Thanks, Chuck

推荐答案

虽然它可能似乎工作。此行取消引用一个空指针：

This code doesn't work, though it may appear to work. This line dereferences a null pointer:

return *static_cast<MyType*>(0);

零，转换为指针类型，然后使用一元 - * 解除引用此空指针。

The zero, cast to a pointer type, results in a null pointer; this null pointer is then dereferenced using the unary-*.

引用空指针导致未定义的行为，因此你的程序可以做任何事情。在你描述的例子中，你得到一个空引用（或者，它看起来你得到一个空引用），但它也是合理的程序崩溃或其他任何发生。

Dereferencing a null pointer results in undefined behavior, so your program may do anything. In the example you describe, you get a "null reference" (or, it appears you get a null reference), but it would also be reasonable for your program to crash or for anything else to happen.

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