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This is a discussion on pointers within the C Programming forums, part of the General Programming Boards category; Originally Posted by std10093 Code: int foo(int array[n]) { //I have access to n?? } Not, in this case, unless ...

  1. #16
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    Quote Originally Posted by std10093 View Post
    Code:
    int foo(int array[n])
    {
      //I have access to n??
    }
    Not, in this case, unless `n' is a compile-time constant (or a macro, that somehow expands to a compile-time constant).
    Right 98% of the time, and don't care about the other 3%.

  2. #17
    SAMARAS std10093's Avatar
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    Quote Originally Posted by c99tutorial View Post
    If you have a three dimensional array declared as `double A[x][y][z]' is it clearer to address an arbitrary element with indices a, b, c as

    A[a][b][c] = foo;

    or

    *(((A+a*x*y)+b*x)+c) = foo;

    ? Especially when there are more than two dimensions, the bottom form becomes tedious and error prone.

    Creating a function that operates on the array A will always require 4 parameters no matter if you use the pointer notation or the array notation: you need to pass one parameter representing the array A, and you need to pass three integer parameters representing the maximum bound of each dimension.

    In other words, your choices for a function bar are pretty much just these two:

    void bar1(int x, int y, int z, double *A);

    void bar2(int x, int y, int z, double A[x][y][z]);
    Wrong. You can not pass *A.

    You pass the array as I said and you handle it normally (of course not as you say...)
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    Quote Originally Posted by std10093 View Post
    You pass the array as I said and you handle it normally (of course not as you say...)
    No. Your example only considers single-dimensional arrays. With an n-dimensional array passed as input into a function, the compiler needs to "know" about the the size of the highest n-1 dimensions of the array. In addition, the programmer needs to know the size of the lowest dimension of the array as well as where the array begins in memory. So in total, you must design your function to accept n+1 parameters. This fact is true independent of whether you use the C99 syntax or C89 syntax. Using the C99 syntax simply lets you use indices inside the function in a natural mathematical way, i.e. "in the normal way".

  4. #19
    SAMARAS std10093's Avatar
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    Yes is the answer. When you want to pass a n dimensional array you pass n stars as pointers and n dimensions. So simple.

    Also, I suppose you understood your mistake before... I guess actually it won't even compile. Tip : Compile the code before you post it

    Enough said and this thread is solved.

    Bye
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    files and sockets come to mind also.

  6. #21
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    Quote Originally Posted by std10093
    When you want to pass a n dimensional array you pass n stars as pointers and n dimensions. So simple.
    Not quite:
    Quote Originally Posted by c99tutorial
    With an n-dimensional array passed as input into a function, the compiler needs to "know" about the the size of the highest n-1 dimensions of the array.
    That is, if you pass an array of arrays as an argument, it is converted to a pointer to an array, not a pointer to a pointer.
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  7. #22
    SAMARAS std10093's Avatar
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    Well, you are correct. I had in mind the dynamic allocation of arrays
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