passing an array to a function (2d array)

This is a discussion on passing an array to a function (2d array) within the C Programming forums, part of the General Programming Boards category; how is it like void function (matrix[][]) and calling it function (matrix[10][10]); ??...

  1. #1
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    passing an array to a function (2d array)

    how
    is it like


    void function (matrix[][])





    and calling it

    function (matrix[10][10]);
    ??

  2. #2
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    try it and see

    but yes its along the lines of

    type arrayname( type [][] )

    standard 2d-array function prototype.
    there are only 10 people in the world, those who know binary and those who dont

  3. #3
    End Of Line Hammer's Avatar
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    More like
    Code:
    #define ROWS 10
    #define COLS 10
    
    void foo(char a[][COLS])
    {
    }
    
    int main(void)
    {
      char myarray[ROWS][COLS];
      foo(myarray);  
      return 0;
    }
    When all else fails, read the instructions.
    If you're posting code, use code tags: [code] /* insert code here */ [/code]

  4. #4
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    why do u only add the number inside the last brackets only?

  5. #5
    ATH0 quzah's Avatar
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    Originally posted by revelation437
    why do u only add the number inside the last brackets only?
    Because that's the way C works. If you specify one dimension, you have to specify every dimension to the right of it. You always have to specify the last one.

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

  6. #6
    End Of Line Hammer's Avatar
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    The compiler needs to know the array size in order to do pointer arithmatic. In this snippet:
    >>void foochar(char name[]);
    the compiler knows that the size of each element in the array is sizeof(char). So when you do name[1], you are adding sizeof(char) to name.

    In this example:
    >>void foochar(char name[][10]);
    the compiler knows the size of name[0][0] is sizeof(char) (as above) and name[0] is size 10*sizeof(char). If you didn't specify the array size in the second brackets, the size equation becomes x*sizeof(char) and your compiler will complain (mine says something like "Size of the type 'signed char[]' is unknown or zero").

    Code:
    #include <stdio.h>
    
    int main(int argc, char *argv[])
    {
      char a[10][10][10];
      printf ("sizeof(a[0][0][0]) = %d\n"
              "sizeof(a[0][0])    = %d\n"
              "sizeof(a[0])       = %d\n",
              sizeof(a[0][0][0]),
              sizeof(a[0][0]),
              sizeof(a[0])
              );
      return(0);
    }
    
    /*
    
    Output:
    
    sizeof(a[0][0][0]) = 1
    sizeof(a[0][0])    = 10
    sizeof(a[0])       = 100
     
     */
    When all else fails, read the instructions.
    If you're posting code, use code tags: [code] /* insert code here */ [/code]

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