Thread: pointers

  1. #1
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    Sep 2009
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    pointers

    what does this statement do *(unsigned short)*a

  2. #2
    Registered User
    Join Date
    Sep 2007
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    1,012
    That's a constraint violation (aka syntax error).

    Presumably you mean:
    Code:
    *(unsigned short *)a;
    This converts a to a pointer to unsigned short and then dereferences it, giving you the value of the unsigned short int at that address.

    This could be useful if a is, say, a pointer to unsigned char/void that is used as a generic pointer to some chunk of memory. You, for one reason or another, know that the pointer actually points to a short (or at the very least you want to treat the value at the address as a short). You can't dereference a directly because it'll either give you a byte (if it's an unsigned char pointer) or it'll be invalid (if it's a void pointer). So you tell the compiler to pretend a is a pointer to an unsigned short, and then dereference it. It's a shorthand for:
    Code:
    unsigned short *s = (unsigned short *)a; /* cast may or may not be necessary */
    *s;

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