解引用空指针 [英] dereferencing the null pointer

问题描述

int* p = 0;
int* q = &*p;

这是未定义的行为或不？我浏览一些相关的问题，但这个具体方面没有显示出来。

Is this undefined behavior or not? I browsed some related questions, but this specific aspect didn't show up.

推荐答案

在这个问题的答案是：这取决于你是以下哪种语言标准： - ）

The answer to this question is: it depends which language standard you are following :-).

在C90和C ++，因为你对空指针进行间接（通过执行 * P ），以及这样做，这不是有效的结果不确定的行为。

In C90 and C++, this is not valid because you perform indirection on the null pointer (by doing *p), and doing so results in undefined behavior.

不过，在C99，这个的是的有效的，良好的，和明确的。在C99，如果一元的操作数​​ - ＆安培; 而获得应用一元的结果 - * 或通过执行下标（ [] ），那么无论是＆安培; 还是 * 或 [] 应用。例如：

However, in C99, this is valid, well-formed, and well-defined. In C99, if the operand of the unary-& was obtained as the result of applying the unary-* or by performing subscripting ([]), then neither the & nor the * or [] is applied. For example:

int* p = 0;
int* q = &*p; // In C99, this is equivalent to int* q = p;

同样，

int* p = 0;
int* q = &p[0]; // In C99, this is equivalent to int* q = p + 0;

从C99§6.5.3.2/ 3：

From C99 §6.5.3.2/3:

如果操作数[的一元＆安培; 运算符]是一元 * 运算符的结果，无论是该运营商也不是＆放大器; 运营商进行评估，其结果是，如果两者都省略了，除了对经营者的约束仍然适用，其结果不是左值

If the operand [of the unary & operator] is the result of a unary * operator, neither that operator nor the & operator is evaluated and the result is as if both were omitted, except that the constraints on the operators still apply and the result is not an lvalue.

同样的，如果操作数是 [] 运算符的结果，无论是＆安培; 运营商，也不是一元 * 由隐含的 [] 进行评估，其结果是，如果在 ＆安培; 运营商被拆除，并在 [] 运营商都改为 + 运营商。

Similarly, if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator were removed and the [] operator were changed to a + operator.

（和它的注脚，＃84）：

(and its footnote, #84):

因此​​，＆放大器; * E 等同于电子（即使电子是一个空指针）

Thus, &*E is equivalent to E (even if E is a null pointer)

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