Giving an argument to function.

This is a discussion on Giving an argument to function. within the C Programming forums, part of the General Programming Boards category; Hi, i have such situation: Code: #include <stdio.h> typedef int x[2][2]; int y(int **i) { i[0][1]=1; return 0; } int ...

  1. #1
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    Giving an argument to function.

    Hi, i have such situation:

    Code:
    #include <stdio.h>
    
    typedef int x[2][2];
    
    int y(int **i)
    {
            i[0][1]=1;
            return 0;
    }
    
    int main()
    {
            x one;
            y(&one);
            printf("%d\n",one[0][1]);
            return 0;
    }
    but i get a warning:
    $ gcc hhhh.c
    hhhh.c: In function `main':
    hhhh.c:15: warning: passing arg 1 of `y' from incompatible pointer type
    and of course, printed value is bad.
    What's wrong with that?
    Regards.

  2. #2
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    Try:
    Code:
            y(one);
    You don't need the address of the double pointer...just the double pointer.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

  3. #3
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    unfortunatelly, without changes...

  4. #4
    and the hat of wrongness Salem's Avatar
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    Having created the typedef, why not use it?

    Code:
    int y ( x i );
    {
            i[0][1]=1;
            return 0;
    }
    int main()
    {
            x one;
            y(one);
            printf("%d\n",one[0][1]);
            return 0;
    }
    > You don't need the address of the double pointer...just the double pointer.
    There are no double pointers.
    There is an array, and a pointer to an array.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  5. #5
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    Thank you. Now it works. I though i have to use pointers to change the object outside the function ( int y(someting *i)).

  6. #6
    ATH0 quzah's Avatar
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    Arrays are not like non-array variables. If you pass an array to a function, it's really passed via reference, so every change inside the function effects the real array. You can't pass arrays by value.

    Quzah.
    Hope is the first step on the road to disappointment.

  7. #7
    and the hat of wrongness Salem's Avatar
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    Just because you've created some typedefs to hide all the 'pointer' bits from sight doesn't mean that C all of a sudden stops passing a pointer to the first element of the array.

    Typedef doesn't add anything new, it just allows you to say more concisely what you could have said another way.
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

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