Pointer for 2D array seems to be 3D in C Post: 302999243

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Top Forums Programming Pointer for 2D array seems to be 3D in C Post 302999243 by yifangt on Thursday 15th of June 2017 07:07:42 PM

06-15-2017

Registered User

Pointer for 2D array seems to be 3D in C

I am struggling with the pointer to 2D-array (cf: 2D array of pointers). Can anybody help me elaborate how the pointer x moves in the memory to access the individual of y[2][6], especially the high lighted lines?
I have talked to one of the curators of the forum, but I am still not quite clear.
Here is my code:

Code:

#include<stdio.h>

int main(void)
{
    int (*x)[2][6];                 //pointer for integers array in size of 2x6 (2 rows x 6 columns),
                                    //.i.e the array is always with size of 12?
//    int (*a[8])[5];                 //Line 9: a is a pointer array of size 8, each for integer array of size 5 

    int y[2][6] = {{11,12,13,14,15,16},
                   {21,22,23,24,25,26}};    //2D array of integers
    int *z;                      //pointer to integer
    int i;

    z = y[0];
    for(i = 0;i<6;i++)
        printf("%d ",z[i]);
    printf("\n");

    x = &y;    // More complicated situation for me!
    x = y;     // Warning: incompatible pointer type.
//    x[0][0] = y[0][0];     // won't work
        printf("   (x[0][0]): %p\n",  x[0][0]);
        printf("  *(x[0][0]): %d\n",*(x[0][0]));        //Q1a
        printf("  x[0][0][0]: %d\n",x[0][0][0]);        //Q1b
        printf("*(x[0][0]+1): %d\n",*(x[0][0]+1));      //Q1c
//        printf("*(x[0][0]+2): %d\n",*(x[0][0]+2));
//        printf("*(x[0][0]+3): %d\n",*(x[0][0]+3));
//        printf("*(x[0][0]+4): %d\n",*(x[0][0]+4));
        printf("*(x[0][0]+5): %d\n",*(x[0][0]+5));

        printf("    x[0][1]: %p\n",   x[0][1]);        //Q2a
        printf("    *x[0][1]: %d\n",*(x[0][1]));
        printf("  x[0][1][0]: %d\n",x[0][1][0]);       //Q2b 
        printf("*(x[0][1]+1): %d\n",*(x[0][1]+1));     //Q2c 
//        printf("*(x[0][1]+2): %d\n",*(x[0][1]+2));
//        printf("*(x[0][1]+3): %d\n",*(x[0][1]+3));
        printf("*(x[0][1]+4): %d\n",*(x[0][1]+4));
        printf("*(x[0][1][4]): %d\n",*(x[0][4]));
        
    printf("&y: %p\n", &y);
    printf(" y: %p\n",  y);
  
    printf(" x: %p\n",  x);
    printf("&x: %p\n", &x);
    
    return 0;
}

1) Although y and &y are the same, but x = y issues warning;
2) Q1a/Q2a is the part I think I understand which is the first element of each row of y.
3) but Q1b/c, and Q2b/c turns out to be 3-D to me.
Can anybody give me a diagram how pointer x moves in the memory for each member of y?
4) Line 9: int (*a[8])[5]; is related, and I put it here for future reference but skip it at this moment.

Thanks a lot!

yifangt

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dxmsvngetdisplayed

DXmSvnGetDisplayed(3X)													    DXmSvnGetDisplayed(3X)

NAME

       DXmSvnGetDisplayed - Retrieves (returns) displayed entry numbers and related information necessary to draw a corresponding display.

SYNOPSIS

       void DXmSvnGetDisplayed(
	       Widget widget,
	       int *entries(),
	       XtPointer *tags(),
	       int *ys(),
	       int len );

PARAMETERS

       The  identifier	(widget  ID)  of the SVN widget.  A pointer to an array of integers to receive the entry numbers of the entries being dis-
       played.	A pointer to an array of longwords to receive the application's entry_tag value  for  each  entry  displayed.	If  tags  are  not
       required,  a  Null  pointer may be passed.  A pointer to an array of longwords to receive the y coordinates for each entry displayed.  If y
       coordinates are not required, a Null pointer can be passed.  The number of entries allocated in the provided array.

DESCRIPTION

       The DXmSvnGetDisplayed routine returns information about the entries that the SVN widget is currently  displaying.   This  information  can
       then  be  used  to keep a simultaneous display up to date with the SVN widget window (in the case of a dialog box, for example, which might
       contain totals for the entries being displayed).

       The application is responsible for managing the memory used to return this list of entries.  As such, note the following: At  the  minimum,
       the  number  of	entries  in the array should be capable of holding at least the number of entries indicated by the value returned from the
       DXmSvnGetNumDisplayed routine.  If there are more entries in the array than the application will need, the SVN widget will  set	the  value
       for those extra entries to 0.  If the capacity of the arrays passed is less than the number of selected entries, only the number of entries
       allocated in the provided array (the value for len) will be returned.

															    DXmSvnGetDisplayed(3X)