引用无效的指针，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

请考虑：

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

* 它与立即& 结合使用，并且没有重载 op& / op * 在玩？

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-end数组元素& myArray [n] ，一个明确等效于& *（myArray + n） c>。>

解决方案<>

/ div>

假设变量`ptr'不包含指向有效对象的指针，如果程序需要对表达式* ptr进行左值到右值转换，则会出现未定义的行为，如[conv .lval]（ISO / IEC 14882：2011，第82页，4.1 [＃1]）。

在& * ptr的评估期间，根据[expr.unary.op]（ISO / IEC 14882：2011，第109页，5.3.1 [＃3]）

$ b，需要对子表达式* ptr进行左值到右值转换
$ b

因此，这是合法的。

Consider:

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

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?

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(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.)

解决方案

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]).

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])

Hence, it is legal.

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