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Home/ Questions/Q 6748353
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T12:33:17+00:00 2026-05-26T12:33:17+00:00

Consider: int* ptr = (int*)0xDEADBEEF; cout << (void*)&*ptr; How illegal is the * ,

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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 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.)

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  1. Editorial Team
    Editorial Team
    2026-05-26T12:33:18+00:00Added an answer on May 26, 2026 at 12:33 pm

    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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