Author : Ted Jensen
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}
printf("\n%d %p %d", row, rowptr[row], rowptr[row]);
if (row > 0)
printf(" %d",(int)(rowptr[row] - rowptr[row-1]));
}
return 0;
}
--------------- End 9.2 ------------------------------------
In the above code rowptr is a pointer to pointer to type int. In this case it points to the first element of an array of pointers to type int. Consider the number of calls to malloc():
To get the array of pointers 1 call
To get space for the rows 5 calls
-----
Total 6 calls
If you choose to use this approach note that while you can use the array notation to access individual elements of the array, e.g. rowptr[row][col] = 17;, it does not mean that the data in the "two dimensional array" is contiguous in memory.
You can, however, use the array notation just as if it were a continuous block of memory. For example, you can write:
rowptr[row][col] = 176;
just as if rowptr were the name of a two dimensional array created at compile time. Of course row and col must be within the bounds of the array you have created, just as with an array created at compile time.
If you want to have a contiguous block of memory dedicated to the storage of the elements in the array you can do it as follows:
Method 4:
In this method we allocate a block of memory to hold the whole array first. We then create an array of pointers to point to each row. Thus even though the array of pointers is being used, the actual array in memory is contiguous. The code looks like this:
----------------- Program 9.3 -----------------------------------
/* Program 9.3 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int **rptr;
int *aptr;
int *testptr;
int k;
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 8; /* or read in at run time */
int row, col;
/* we now allocate the memory for the array */
aptr = malloc(nrows * ncols * sizeof(int));
if (aptr == NULL)
{
puts("\nFailure to allocate room for the array");
exit(0);
}
/* next we allocate room for the pointers to the rows */
rptr = malloc(nrows * sizeof(int *));
if (rptr == NULL)
{
puts("\nFailure to allocate room for pointers");
exit(0);
}
/* and now we 'point' the pointers */
for (k = 0; k < nrows; k++)
{
rptr[k] = aptr + (k * ncols);
}
/* Now we illustrate how the row pointers are incremented */
printf("\n\nIllustrating how row pointers are incremented");
printf("\n\nIndex Pointer(hex) Diff.(dec)");
for (row = 0; row < nrows; row++)
{
printf("\n%d %p", row, rptr[row]);
if (row > 0)
printf(" %d",(rptr[row] - rptr[row-1]));
}
printf("\n\nAnd now we print out the array\n");
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
rptr[row][col] = row + col;
printf("%d ", rptr[row][col]);
}
putchar('\n');
}
puts("\n");
/* and here we illustrate that we are, in fact, dealing with
a 2 dimensional array in a contiguous block of memory. */
printf("And now we demonstrate that they are contiguous in memory\n");
testptr = aptr;
for (row = 0; row < nrows; row++)
{
for (col = 0; col < ncols; col++)
{
printf("%d ", *(testptr++));
}
putchar('\n');
}
return 0;
}
------------- End Program 9.3 -----------------
Consider again, the number of calls to malloc()
To get room for the array itself 1 call
To get room for the array of ptrs 1 call
----
Total 2 calls
Now, each call to malloc() creates additional space overhead since malloc() is generally implemented by the operating system forming a linked list which contains data concerning the size of the block. But, more importantly, with large arrays (several hundred rows) keeping track of what needs to be freed when the time comes can be more cumbersome. This, combined with the contiguousness of the data block that permits initialization to all zeroes using memset() would seem to make the second alternative the preferred one.
As a final example on multidimensional arrays we will illustrate the dynamic allocation of a three dimensional array. This example will illustrate one more thing to watch when doing this kind of allocation. For reasons cited above we will use the approach outlined in alternative two. Consider the following code:
------------------- Program 9.4 -------------------------------------
/* Program 9.4 from PTRTUT10.HTM 6/13/97 */
#include <stdio.h>
#include <stdlib.h>
#include <stddef.h>
int X_DIM=16;
int Y_DIM=5;
int Z_DIM=3;
int main(void)
{
char *space;
char ***Arr3D;
int y, z;
ptrdiff_t diff;
/* first we set aside space for the array itself */
space = malloc(X_DIM * Y_DIM * Z_DIM * sizeof(char));
/* next we allocate space of an array of pointers, each
to eventually point to the first element of a
2 dimensional array of pointers to pointers */
Arr3D = malloc(Z_DIM * sizeof(char **));
/* and for each of these we assign a pointer to a newly
allocated array of pointers to a row */
for (z = 0; z < Z_DIM; z++)
{
Arr3D[z] = malloc(Y_DIM * sizeof(char *));
/* and for each space in this array we put a pointer to
the first element of each row in the array space
originally allocated */
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