取消引用空指针未定义行为。

未定义行为意味着任何事情都可能发生，因此无法定义行为。

An Undefined Behavior means anything can happen, So it is not possible to define a behavior for this.

不确定的是，我将第n次添加这个C ++标准报价，但似乎需要。

Admittedly, I am going to add this C++ standard quote for the nth time, but seems it needs to be.

关于未定义的行为，

C ++标准版1.3.24说明：

C++ Standard section 1.3.24 states:

允许未定义的行为范围该情况完全具有不可预测的结果，在以文件化的环境特征（有或没有诊断消息的发布）的方式执行翻译或程序期间行为，终止翻译或执行（发出诊断消息）。

Permissible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).

注意：

此外，请注意： br>
使用返回的引用或指向函数内的局部变量的指针也是未定义的行为。你应该使用new分配指针在freestore（堆），然后返回一个引用/指针。

NOTE:
Also, just to bring it to your notice:
Using a returned reference or pointer to a local variable inside a function is also an Undefined Behavior. You should be allocating the pointer on freestore(heap) using new and then returning a reference/pointer to it.

编辑：

正如@James McNellis在评论中恰当地指出，

如果未使用返回的指针或引用，则行为明确。