取消引用无效指针，然后获取结果的地址 [英] Dereferencing an invalid pointer, then taking the address of the result

问题描述

考虑:

int* ptr = (int*)0xDEADBEEF;
cout << (void*)&*ptr;

* 有多非法，因为它与直接的 & 结合使用，并且没有重载的 op&/op* 在玩?

How illegal is the *, given that it's used in conjunction with an immediate & and given that there are no overloaded op&/op* in play?

(这对于寻址过去结束的数组元素 &myArray[n] 具有特殊的影响，该表达式明确等效于 &*(myArray+n). 这个问答 解决了更广泛的问题，但我觉得它并没有真正满足上述问题.)

(This has particular ramifications for addressing a past-the-end array element &myArray[n], an expression which is explicitly equivalent to &*(myArray+n). This Q&A addresses the wider case but I don't feel that it ever really satisfied the above question.)

推荐答案

假设变量 `ptr' 不包含指向有效对象的指针，如果程序需要将表达式 `*ptr'，在 [conv.lval](ISO/IEC 14882:2011，第 82 页，4.1 [#1])中指定.

Assuming the variable `ptr' does not contain a pointer to a valid object, the undefined behavior occurs if the program necessitates the lvalue-to-rvalue conversion of the expression `*ptr', as specified in [conv.lval] (ISO/IEC 14882:2011, page 82, 4.1 [#1]).

根据 [expr.unary.op] (ISO/IEC 14882:2011,第 109 页，5.3.1 [#3])

During the evaluation of `&*ptr' the program does not necessitate the lvalue-to-rvalue conversion of the subexpression `*ptr', according to [expr.unary.op] (ISO/IEC 14882:2011, page 109, 5.3.1 [#3])

因此，这是合法的.

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