function reveives a 2-d array as parameter

This is a discussion on function reveives a 2-d array as parameter within the C Programming forums, part of the General Programming Boards category; hi all, how can i write a function which receives a 2-d array? my version: Code: #include <stdio.h> int main(void) ...

  1. #1
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    function reveives a 2-d array as parameter

    hi all,
    how can i write a function which receives a 2-d array?
    my version:
    Code:
    #include <stdio.h>
    int main(void)
    {
      void fun(int *arr[10]);
    
      int twodarray[10][10] = {0};
    
      fun(twodarray);
    
    return 0;
    }
    
     void fun(int *arr[10])
     {
       int i;
       for(i=0; i<10; i++)
       {
         printf("%d", arr[0][i]);
       }
      }
    it can't be compiled, here's error mesaage
    Code:
    9 D:\Programes\OfficeWork\Dev-Cpp\Fun2darray\main.cpp cannot convert `int (*)[10]' to `int**' for argument `1' to `void fun(int**)'
    Code:
    void fun(int **arr)
    {....}
    not working neither
    BTW, how can i deal with a 3-d array?

    many thanks!

  2. #2
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    Code:
    fun(int arr[][10]) ...
    For a 3D array, you will need to specify:
    Code:
    fun3d(int arr[][10][10]);
    Note that just becase a one-dimensional array will convert to a pointer, it doesn't automatically make a 2D array into a pointer to pointer - the compiler needs to know the dimensions aside from the first one [how LONG the 2D array is].

    --
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    Compilers can produce warnings - make the compiler programmers happy: Use them!
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  3. #3
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    great thanks,
    i try this and it seems work fine, but i don't know why,
    can u help to explain it?
    Code:
     void fun(int (*arr)[10])

  4. #4
    Jack of many languages Dino's Avatar
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    Thanks Mats.

    Code:
    #include <stdio.h>
    
    void fun(int arr[][10]);
    
    int main(void) { 
    	int i, j, k = 0; 
    	
    	int twodarray[10][10] = {0};
    	
    	for (i = 0 ; i < 10 ; i++) { 
    		for (j = 0 ; j < 10 ; j++ ) { 
    			twodarray[i][j] = k ; 
    			k++ ;
    		}
    	}
    	fun(twodarray);
    
    return 0;
    }
    
    void fun(int arr[][10]) {
    	int i, j ;
    	for(i=0; i<10; i++) {
    		for ( j = 0 ; j < 10 ; j++ ) {
    			printf("&#37;d:%d = %d\n", i, j, arr[i][j]);
    		}
    	}
    }
    Todd

  5. #5
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    Yes, that's another variation. It basicly says "this is a pointer to an array of ten elements" [as opposed to the original form of "10 pointers to integers" - which isn't what you want].

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  6. #6
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    thanks again,
    i've another question now,
    how to write a function that returns an address of a 2-d array?
    mine: (revised twice - thanks to matsp)
    Code:
    #include <stdio.h>
    void fun(int (*arr)[10]);
    int send(void);
    
    int main(void)
    {
        int (*ptr)[10];
        ptr = send();
        fun(ptr);
    return 0;
    }
     void fun(int (*arr)[10])
     {
       int i;
       for(i=0; i<10; i++)
       {
         printf("&#37;d", arr[0][i]);
       }
      }
      
    int send(void)        //here is the problem.....
    {
        int static twodarray[10][10] = {0};
        return twodarray;  
    }
    (thank you Todd, but pleaseforgive me using my version ...for simplicity
    Last edited by albert3721; 12-04-2007 at 06:44 AM. Reason: optimize-reedit-owe to matsp

  7. #7
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    im propably using your first code.......... now i think so this would work for you........ im printing array[9][0]...........array[9][9].....

    Code:
    #include <stdio.h>
    int main(void)
    {
      int ** fun(int **arr);
      int **ptr;
      int i;
      int twodarray[10][10] = {0};
    
      ptr=fun(twodarray);
      for(i=0; i<10; i++)
       {
         printf("&#37;d\n",* ((ptr+9)+i));
       }
    
    return 0;
    }
    
     int ** fun(int **arr)
     {
       int i;
       for(i=0; i<10; i++)
       {
         printf("%d\n",* ((arr+9)+i));
       }
       return arr;
     }

  8. #8
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    Quote Originally Posted by albert3721 View Post
    thanks again,
    i've another question now,
    how to write a function that returns an address of a 2-d array?
    mine:
    Code:
    #include <stdio.h>
    void fun(int (*arr)[10]);
    int send(void);
    
    int main(void)
    {
        int (*ptr)[10];
        ptr = send;
        fun(ptr);
    return 0;
    }
     void fun(int (*arr)[10])
     {
       int i;
       for(i=0; i<10; i++)
       {
         printf("%d", arr[0][i]);
       }
      }
      
    int send(void)        //here is the problem.....
    {
        int twodarray[10][10] = {0};
        return twodarray;  
    }
    (thank you Todd, but pleaseforgive me using my version ...for simplicity
    1. You are not actually calling send, you are assigning a pointer with the address of the function send.
    2. You are returning a local variable [twodarray in send], so that variable will "disappear" when the funciton returns.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  9. #9
    Jack of many languages Dino's Avatar
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    OK, this compiles, but crashes in the first assignment in red. Something is wrong.

    Code:
    #include <stdio.h>
    #include <stdlib.h> 
    
    int ** fun1(void);
    void fun2(int **) ; 
    
    int main(void) { 
    	int i, j, k = 0; 
    	
    	int ** twodarray ;  
    			
    	twodarray = fun1();
    	for (i = 0 ; i < 10 ; i++) { 
    		for (j = 0 ; j < 10 ; j++ ) { 
    			twodarray[i][j] = k ; 
    			k++ ;
    		}
    	}
    	fun2(twodarray) ; 
    	return 0;
    }
    
    int ** fun1(void) {
    	int **arr ;  
    	arr = calloc( 10*10 , sizeof(int) ) ; 
    	return arr ; 
    	
    }
    
    void fun2(int **arr) {
    	int i, j ; 
    	for( i = 0 ; i < 10 ; i++ ) {
    		for ( j = 0 ; j < 10 ; j++ ) {
    			printf("&#37;d:%d = %d\n", i, j, arr[i][j]);
    		}
    	}
    }

  10. #10
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    Yes, yet again, we're attempting to convert ** into a 2D array. That's not what ** means. ** means that you have ONE variable that contains a pointer to a pointer to a integer.

    A two dimensional array is a block of memory that is x elements wide and y elements "high".

    You can use an array of y pointers that point to arrays of x elements, and a pointer to pointer that points to the y pointers. Something like this:

    Code:
    int ** fun1(void) {
    	int **arr ;  
            arr = calloc(10, sizeof(int *));   // 10 pointers to integer. 
            for(i = 0; i < 10; i++) 
            {
    	   arr[i] = calloc( 10 , sizeof(int) ) ; // 10 integers. 
            }
    	return arr ; 
    }
    Just don't forget to do the opposite to free them later on.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  11. #11
    Jack of many languages Dino's Avatar
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    Got it. Thanks again!

    Code:
    #include <stdio.h>
    #include <stdlib.h>
    
    int ** fun1(void);
    void fun2(int **) ; 
    void fun3(int **) ; 
    
    int main(void) { 
    	int i, j, k = 0; 
    	
    	int ** twodarray ;  
    			
    	twodarray = fun1();            // allocate the memory 
    	for (i = 0 ; i < 10 ; i++) {   // assign values 
    		for (j = 0 ; j < 10 ; j++ ) { 
    			twodarray[i][j] = k ; 
    			k++ ;
    		}
    	}
    	fun2(twodarray) ;   // print out the values
    	fun3(twodarray) ;   // free the memory 
    	return 0;
    }
    
    int ** fun1(void) {
    	int i,  **arr ;  
    	arr = calloc( 10 , sizeof(int) ) ; 
    		printf("Heap allocated... address=&#37;p\n", arr) ; 
    	for (i = 0 ; i < 10 ; i++) { 
    		arr[i] = calloc(10, sizeof(int) ) ; 
    		printf("Heap allocated... address=%p\n", arr[i]) ; 
    	}
    	return arr ; 
    	
    }
    
    void fun2(int **arr) {
    	int i, j ; 
    	for( i = 0 ; i < 10 ; i++ ) {
    		for ( j = 0 ; j < 10 ; j++ ) {
    			printf("%d:%d = %d\n", i, j, arr[i][j]);
    		}
    	}
    }
    
    void fun3(int ** arr) {
    	int i ;   
    	for (i = 0 ; i < 10 ; i++) { 
    		printf("Freeing Heap... address=%p\n", arr[i]) ; 
    		free(arr[i]) ; 
    	}
    	printf("Freeing Heap... address=%p\n", arr) ; 
    	free(arr) ; 
    }

  12. #12
    Jack of many languages Dino's Avatar
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    I think my confusion was that the compiler was doing a lot of work for me behind the scenes when I define

    twodarray[10][10].

    I envisioned this as a block of 400 bytes (assuming an int is 4 bytes), but that is obviously not the case. The compiler is (essentially) hard coding the pointers for me, so there is really:

    40 bytes for the 10 pointers ("i")
    10 * 40 bytes for all the 2nd dimension pointers ("j")
    400 bytes for the actual integers.

    So, 840 bytes for an array of 100 integers. Is this correct?

    Todd

  13. #13
    and the hat of wrongness Salem's Avatar
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    You can return a pointer to a true 2D array inside a function, providing that the array is declared static (inside the function).

    However, the syntax for the function declaration is a sight to behold!
    Code:
    #include <stdio.h>
    
    int (*func(void))[10] {
        static int arr[10][10] = {
            { 1 },
            { 2 },
        };
        return arr;
    }
    
    int main ( ) {
        int r, c;
        int (*arr)[10] = func();
        for ( r = 0 ; r < 10 ; r++ ) {
            for ( c = 0 ; c < 10 ; c++ ) {
                printf( "&#37;d ", arr[r][c] );
            }
            printf( "\n" );
        }
        return 0;
    }
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  14. #14
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    Quote Originally Posted by Salem View Post
    However, the syntax for the function declaration is a sight to behold!
    "Beauty is in the eye of the beholder" as they say - in this case, I don't like it at all.

    Also, a static return array is only good if you either:
    - Only ever call it for ONE array at any given time - otherwise, the second array is an alias of the first one.
    - Definitely DON'T use multithreading.

    There are several ways to manage 2D arrays that is MUCH easier than returning an array from a function. For example, you can wrap the array into a struct, and use a pointer to the struct.

    Or pass a 2D array in from the lower level call, rather than return it from a higher call.

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    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

  15. #15
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    I envisioned this as a block of 400 bytes (assuming an int is 4 bytes), but that is obviously not the case. The compiler is (essentially) hard coding the pointers for me, so there is really:

    40 bytes for the 10 pointers ("i")
    10 * 40 bytes for all the 2nd dimension pointers ("j")
    400 bytes for the actual integers.
    Yes, a simple 2D array is a block of 400 bytes. There are no "hidden pointers", etc when the compiler deals with it. The pointers become a must when you try to create your own 2D array, and use it as a pointer to pointer. [This makes it flexible as to what the two dimensions are too - you just change the internals of the function to make it 20 x 20].

    You can ALSO create a block of 400 bytes, and return it using Salems "something to behold" syntax:
    Code:
    int (*func(void))[10] {
        int (*arr)[10];
        arr = calloc(10 * 10, sizeof(int));
        return arr;
    }
    This code only works for a [x][10] array - as it's a pointer to array of 10 integers.

    --
    Mats
    Compilers can produce warnings - make the compiler programmers happy: Use them!
    Please don't PM me for help - and no, I don't do help over instant messengers.

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