在C中是否定义了& * NULL? [英] Is &*NULL well-defined in C?
问题描述
在以下哪个版本的C标准(如果有)中定义明确?
In what version(s) of the C standards (if any) is the following well-defined?
void foo(void) {
char *nullPtr = NULL;
&*nullPtr;
}
请注意，我没有将结果分配给任何内容-第二行是一个简单的语句.
Note that I am not assigning the result to anything - the second line is a simple statement.
这个应该是一个带有明显答案的问题，但是(似乎在此类问题上经常发生)我听到很多人说答案显然没有定义"为明显"定义".
This should be a question with an obvious answer, but (as seemingly happens way too often on such questions) I have heard just as many people say the answer is "obviously undefined" as "obviously defined".
相对来说，以下内容如何? foo是否应该读取c?
On a rather related note, what about the following? Should foo produce a read of c?
extern volatile char c;
void bar(void) {
volatile char *nonnullptr = &c;
&*nonnullptr;
}
(同一问题的C ++版本:在C ++中是否定义了& * NULL?)
(C++ version of the same question: Is &*NULL well-defined in C++?)
推荐答案
尽管尝试取消引用空指针会导致未定义的行为，所以*nullPtr是非法的，而&*nullPtr则是定义明确的. 根据C11标准草案的脚注102 :
While attempts to dereference a null pointer cause undefined behavior, so *nullPtr is illegal, &*nullPtr is perfectly well-defined. According to footnote 102 in the C11 Draft Standard:
因此，& * E 等同于 E (即使 E 是空指针)，....
Thus, &*E is equivalent to E (even if E is a null pointer),....
这是因为对于一元&运算符(§6.5.3.2¶3):
This is a result of the fact that, for the unary & operator (§6.5.3.2 ¶3):
如果操作数是一元 * 运算符的结果，则不会对该运算符和& 运算符求值，并且结果好像都被省略了一样. ...
If the operand is the result of a unary * operator, neither that operator nor the & operator is evaluated and the result is as if both were omitted,....
C99标准具有相同的语言，但是在C90标准中没有出现，我对该标准的理解是&*nullPtr的确会在C99之前的实现中引起不确定的行为.
The C99 Standard has the same language, but this does not appear in the C90 Standard, and my reading of that standard is that &*nullPtr would indeed cause undefined behavior in pre-C99 implementations.
根据C90标准(第6.3.2.3节):
From the C90 Standard (§6.3.2.3):
一元& (地址)运算符的结果是指向由其操作数指定的对象或函数的指针.
The result of the unary & (address-of) operator is a pointer to the object or function designated by its operand....
和:
一元*运算符表示间接....如果为指针分配了无效值，则一元*运算符的行为是不确定的.
The unary * operator denotes indirection.... If an invalid value has been assigned to the pointer, the behavior of the unary * operator is undefined.
Curiously, I don't see any discussion of this change in the C99 Rationale, though I may just be not finding it.
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