Pointer question

This is a discussion on Pointer question within the C++ Programming forums, part of the General Programming Boards category; Hey all, If I have a pointer: SomeObject *pointer; and I want to access a method from that objects class, ...

  1. #1
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    Pointer question

    Hey all,

    If I have a pointer:
    SomeObject *pointer;

    and I want to access a method from that objects class, I would do it via pointer->someMethod.

    However if I say that pointer = new SomeObject[numelements]; then when I want to do pointer[index]->someMethod it fails but pointer[index].someMethod works. I'm new to C++ so I'm not sure why this is the case.

    Thanks
    Last edited by jcafaro10; 03-04-2009 at 06:16 PM.

  2. #2
    Kiss the monkey. CodeMonkey's Avatar
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    [] dereferences (did I spell that right?). That means [] takes pointers to what-they-point-to.

    ptr+1 is a pointer.
    *(ptr+1) is some type.
    ptr[1] is the same as *(ptr+1).

    we use -> for pointers and . for regular stuff.
    "If you tell the truth, you don't have to remember anything"
    -Mark Twain

  3. #3
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    pointer[index]->someMethod is equal to *(pointer[index]).someMethod but since pointer[index] is an array you dont need the "*" sign.
    If it wasnt an array like *(pointer).someMethod, this pointer->someMethod would work.

    An array is already like a pointer thats why you dont need the dereferencing sign.

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