其中哪些将创建一个空指针？ [英] Which of these will create a null pointer?

问题描述

标准说，取消引用空指针会导致未定义的行为。但是什么是空指针？在下面的代码中，我们称之为空指针：

The standard says that dereferencing the null pointer leads to undefined behaviour. But what is "the null pointer"? In the following code, what we call "the null pointer":

struct X
{
  static X* get() { return reinterpret_cast<X*>(1); }
  void f() { }
};

int main()
{
  X* x = 0;
  (*x).f(); // the null pointer?  (1)

  x = X::get();
  (*x).f(); // the null pointer?  (2)

  x = reinterpret_cast<X*>( X::get() - X::get() );
  (*x).f(); // the null pointer?  (3)

  (*(X*)0).f(); // I think that this the only null pointer here (4)
}

是空指针的解引用仅在最后一种情况下发生。我对吗？根据C ++标准，编译时空指针和运行时之间是否有区别？

My thought is that dereferencing of the null pointer takes place only in the last case. Am I right? Is there difference between compile time null pointers and runtime according to C++ Standard?

推荐答案

只有第一个和最后一个是空指针。其他是 reinterpret_cast 的结果，因此对实现定义的指针值进行操作。它们的行为是否未定义取决于在所转换的地址是否有对象。

Only the first and the last are null pointers. The others are results of reinterpret_cast and thus operate on implementation defined pointer values. Whether the behavior is undefined for them depends on whether there is an object at the address you casted to.

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