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?
(This has part开发者_StackOverflow社区icular 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.)
According to the specification, the effect of dereferencing an invalid pointer itself produces undefined behaviour. It doesn't matter what you do after dereferencing it.
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.
It is legal. Why wouldn't it be? You're just setting a value to a pointer, and then accessing to it. However, assigning the value by hand must be obviously specified as undefined behavior, but that's the most a general specification can say. Then, you use it in some embedded software controller, and it will give you the correct memory-mapped value for some device...
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