Thread: Confused with pointers and pass by reference

  1. #16
    SAMARAS std10093's Avatar
    Join Date
    Jan 2011
    Nice, France
    Quote Originally Posted by AndiPersti View Post
    Basically *&i == i, because & and * cancel.
    That was my point at the above post!

    About tou example i think i got it.But let's check to be sure
    #include <stdio.h>
    int main(void)
        int i, *ptr1, **ptr2, ***ptr3;
        i = 10;
        ptr1 = &i;
        ptr2 = &ptr1;
        ptr3 = &ptr2;
        printf("%d\n", i == **ptr2);
        printf("%d\n", i == **&ptr1);
        printf("%d\n", *&i == ***ptr3);
        printf("%d\n", &i == **&*&ptr2);
        return 0;
    ptr2 is a 'double' pointer so because of line 9 points to ptr1 and then points(finally) to i.So line 12 will output 1.
    Line 13 * cancels & so we have i==*ptr1,of course true,so output 1.
    Line 14 is equivalent to i==***ptr3 which is true,becaues ptr3 is a 'triple pointer' which first points to ptr2(line 9),then points to ptr1 and finally points to i(the value of i).So output 1.
    Line 15 is equivalent to &i==*ptr2 .ptr2 is a double pointer,so it points first to ptr1 and finally at the value 1.So *ptr2 is going to give us the value of ptr1,which is the adress of i and so output is going to be again 1.

  2. #17
    Registered User
    Join Date
    May 2012
    Quote Originally Posted by std10093 View Post
    By writing
    it is like 'killing' the pointer
    Remember that & 'kills' the *
    For me "killing" means to get rid of something (it's destructive).
    When you add the & operator to a function argument, you don't get rid of the pointer but you construct a pointer to that argument (it's constructive).

    That's why I think "killing" is misleading.

    Bye, Andreas

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