Thread: why use the -> operator?

  1. #1
    Registered User
    Join Date
    Jun 2004

    why use the -> operator?

    I was just wondering why, when using object pointers, you must use the -> operatior. Why couldnīt you simply follow the standard method like: *p.funct() if you were accesing the funct() member through the *p pointer?

    Thanks for the help

  2. #2
    Guest Sebastiani's Avatar
    Join Date
    Aug 2001
    Waterloo, Texas
    sure, you can do that - but because of operator precedence you'll need to put a parenthesis around the pointer dereference, ie: (*p).funct()
    #include <cmath>
    #include <complex>
    bool euler_flip(bool value)
        return std::pow
            std::complex<float>(0, 1) 
            * std::complex<float>(std::atan(1.0)
            *(1 << (value + 2)))
        ).real() < 0;

  3. #3
    Super Moderator
    Join Date
    Sep 2001
    And it's a lot easier to just write -> than to write all the parentheses and *'s, let alone site there and figure out where they're all supposed to go.

  4. #4

    For example sake, let's create a structure called point:
    struct point {
    	int x;
    	int y;
    The following declaration says that p is a pointer to a structure of type struct point. If p points to a point structure, *p is the structure, and (*p).x and (*p).y are the members:
    struct point *p;
    To use p, we might write, for example:
    struct point *p;
    printf("(%d %d)\n", (*p).x, (*p).y);
    The parenthesis are necessary in (*p).x because the precedence of the structure member operator (.) is higher than (*). The expression *p.x means *(p).x, which is illegal here because (x) is not a pointer.

    Pointers to structures are so frequently used that an alternative notation is provided as a shorthand. If (p) is a pointer to a structure, then the following code refers to the particular member:

    Note: Both (.) and (->) associate from left to right.

    Edit: Takes me forever to post. Heh.

    - Stack Overflow
    Segmentation Fault: I am an error in which a running program attempts to access memory not allocated to it and core dumps with a segmentation violation error. This is often caused by improper usage of pointers, attempts to access a non-existent or read-only physical memory address, re-use of memory if freed within the same scope, de-referencing a null pointer, or (in C) inadvertently using a non-pointer variable as a pointer.

  5. #5
    Carnivore ('-'v) Hunter2's Avatar
    Join Date
    May 2002
    Also, when you get into operator overloading, * and -> often will not mean "dereference" any more; rather, they will mean "dereference one of my member variables", in which case I personally find -> to be more intuitive - though I imagine that's totally up to you.
    Just Google It. √

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