Pointers and Dynamic Allocation of Memory
Pointers and Dynamic Allocation of Memory There are times when it is convenient to allocate memory at
run time using malloc(), calloc(), or other allocation functions.
Using this approach permits postponing the decision on the size
of the memory block need to store an array, for example, until
run time. Or it permits using a section of memory for the
storage of an array of integers at one point in time, and then
when that memory is no longer needed it can be freed up for other
uses, such as the storage of an array of structures.
When memory is allocated, the allocating function (such as
malloc(), calloc(), etc.) returns a pointer. The type of this
pointer depends on whether you are using an older K&R compiler or
the newer ANSI type compiler. With the older compiler the type
of the returned pointer is char, with the ANSI compiler it is
void.
If you are using an older compiler, and you want to allocate
memory for an array of integers you will have to cast the char
pointer returned to an integer pointer. For example, to allocate
space for 10 integers we might write:
int *iptr;
iptr = (int *)malloc(10 * sizeof(int));
if(iptr == NULL)
{ .. ERROR ROUTINE GOES HERE .. }
If you are using an ANSI compliant compiler, malloc() returns
a void pointer and since a void pointer can be assigned to a
pointer variable of any object type, the (int *) cast shown above
is not needed. The array dimension can be determined at run time
and is not needed at compile time. That is, the “10″ above could
be a variable read in from a data file or keyboard, or calculated
based on some need, at run time.
Because of the equivalence between array and pointer
notation, once iptr has been assigned as above, one can use the
array notation. For example, one could write:
int k;
for(k = 0; k < 10; k++
iptr[k] = 2;
to set the values of all elements to 2.
Even with a reasonably good understanding of pointers and
arrays, one place the newcomer to C is likely to stumble at first
is in the dynamic allocation of multi-dimensional arrays. In
general, we would like to be able to access elements of such
arrays using array notation, not pointer notation, wherever
possible. Depending on the application we may or may not know
both dimensions at compile time. This leads to a variety of ways
to go about our task.
As we have seen, when dynamically allocating a one
dimensional array the dimension can be determined at run time.
Now, when using dynamic allocation of higher order arrays, we
never need to know the first dimension at compile time. Whether
we need to know the higher dimensions depends on how we go about
writing the code. Here I will discuss various methods of
dynamically allocating room for 2 dimensional arrays of integers.
First we will consider cases where the 2nd dimension is known
at compile time.
METHOD 1:
One way of dealing with the problem is through the use of the
“typedef” keyword. To allocate a 2 dimensional array of integers
recall that the following two notations result in the same object
code being generated:
multi[row][col] = 1; *(*(multi + row) + col) = 1;
It is also true that the following two notations generate the
same code:
multi[row] *(multi + row)
Since the one on the right must evaluate to a pointer, the
array notation on the left must also evaluate to a pointer. In
fact multi[0] will return a pointer to the first integer in the
first row, multi[1] a pointer to the first integer of the second
row, etc. Actually, multi[n] evaluates to a pointer to that
array of integers which makes up the n-th row of our 2
dimensional array. That is, multi can be thought of as an array
of arrays and multi[n] as a pointer to the n-th array of this
array of arrays. Here the word “pointer” is being used
to represent an address value. While such usage is common in the
literature, when reading such statements one must be careful to
distinguish between the constant address of an array and a
variable pointer which is a data object in itself.
Consider now:
———————————————–
#include <stdio.h>
#define COLS 5
typedef int RowArray[COLS];
RowArray *rptr;
int main(void)
{
int nrows = 10;
int row, col;
rptr = malloc(nrows * COLS * sizeof(int))
for(row = 0; row < nrows; row++)
for(col = 0; col < COLS; col++)
{
rptr[row][col] = 17;
}
}
————————————————-
Here I have assumed an ANSI compiler so a cast on the void
pointer returned by malloc() is not required. If you are using
an older K&R compiler you will have to cast using:
rptr = (RowArray *)malloc(…. etc.
Using this approach, “rptr” has all the characteristics of an
array name and array notation may be used throughout the rest of
the program. That also means that if you intend to write a
function to modify the array contents, you must use COLS as a
part of the formal parameter in that function, just as we did
when discussing the passing of two dimensional arrays to a
function.
METHOD 2:
In the METHOD 1 above, rptr turned out to be a pointer to
type “one dimensional array of COLS integers”. It turns out that
there is syntax which can be used for this type without the need
of typedef. If we write:
int (char *xptr)[COLS];
the variable xptr will have all the same characteristics as the
variable rptr in METHOD 1 above, and we need not use the
“typedef” keyword. Here xptr is a pointer to an array of
integers and the size of that array is given by the #defined
COLS. The parenthesis placement makes the pointer notation
predominate, even though the array notation has higher
precedence. i.e. had we written
int char *xptr[COLS];
we would have defined xptr as an array of pointers holding the
number of pointers equal to that #defined by COLS. Which is not
the same thing at all. However, arrays of pointers have their
use in the dynamic allocation of two dimensional arrays, as will
be seen in the next 2 methods.
METHOD 3:
Consider the case where we do not know the number of elements
in each row at compile time, i.e. both the number of rows and
number of columns must be determined at run time. One way of
doing this would be to create an array of pointers to type int
and then allocate space for each row and point these pointers at
each row. Consider:
——————————————————
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 10; /* or read in at run time */
int row, col;
int **rowptr;
rowptr = malloc(nrows * sizeof(int *));
if(rowptr == NULL)
{
puts(”\nFailure to allocate room for row pointers.\n”);
exit(0);
}
printf(”\n\n\nIndex Pointer(hex) Pointer(dec) Diff.(dec)”);
for(row = 0; row < nrows; row++)
{
rowptr[row] = malloc(ncols * sizeof(int));
if(rowptr[row] == NULL)
{
printf(”\nFailure to allocate for row[%d]\n”,row);
exit(0);
}
printf(”\n%d %p %d”, row, rowptr[row], rowptr[row]);
if(row > 0)
printf(” %d”,(int)(rowptr[row] - rowptr[row-1]));
}
return 0;
}
—————————————————
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.
But, you can 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 it is desired to have a contiguous block of memory
dedicated to the storage of the elements in the array it can be
done 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:
———————————————————-
#include <stdio.h>
#include <stdlib.h>
#include <conio.h>
int main(void)
{
int **rptr;
int *aptr;
int *testptr;
int k;
int nrows = 5; /* Both nrows and ncols could be evaluated */
int ncols = 10; /* 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 */
clrscr();
for(k = 0; k < nrows; k++)
{
rptr[k] = aptr + (k * ncols);
}
printf(”\n\n\nIndex Pointer(hex) Pointer(dec) Diff.(dec)”);
for(row = 0; row < nrows; row++)
{
printf(”\n%d %p %d”, row, rptr[row], rptr[row]);
if(row > 0)
printf(” %d”,(int)(rptr[row] - rptr[row-1]));
}
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\n\n”);
/* and here we illustrate that we are, in fact, dealing with
a 2 dimensional array in a _contiguous_ block of memory. */
testptr = aptr;
for(row = 0; row < nrows; row++)
{
for(col = 0; col < ncols; col++)
{
printf(”%d “, *(testptr++));
}
putchar(’\n’);
}
return 0;
}
——————————
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 he
contiguousness of the data block which 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:
—————————————————————–
#include <stdio.h>
#include <stdlib.h>
#include <mem.h>
#include <conio.h>
int X_DIM=16;
int Y_DIM=8;
int Z_DIM=4;
int main(void)
{
char ***space;
char ***Arr3D;
int x, 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 */
for(y = 0; y < Y_DIM; y++)
{
Arr3D[z][y] = ((char *)space + (z*(X_DIM * Y_DIM) + y*X_DIM));
}
}
/* And, now we check each address in our 3D array to see if
the indexing of the Arr3d pointer leads through in a
continuous manner */
for(z = 0; z < Z_DIM; z++)
{
printf(”Location of array %d is %p\n”, z, *Arr3D[z]);
for( y = 0; y < Y_DIM; y++)
{
printf(” Array %d and Row %d starts at %p”, z, y, Arr3D[z][y]);
diff = Arr3D[z][y] - (char *)space;
printf(” diff = %d “,diff);
printf(” z = %d y = %d\n”, z, y);
}
getch();
}
return 0;
}
———————————————————–
If you have followed this tutorial up to this point you
should have no problem deciphering the above on the basis of the
comments alone. There is one line that deserves a bit of special
attention however. It reads:
Arr3D[z][y] = ((char *)space + (z*(X_DIM * Y_DIM) + y*X_DIM));
Note that here “space” is cast to a character pointer, which
is the same type as Arr3D[z][y]. A thing to be careful of,
however, is where that cast is made. If the cast were made
outside the overall parenthesis as in…
Arr3D[z][y] = (char *)(space + (z*(X_DIM * Y_DIM) + y*X_DIM));
the code fails. The reason is that the cast, in this case, is
not so much to make the types on each side of the assignment
operator match, as it is to make the pointer arithmetic work.
Recall that when dealing with pointer arithmetic in something
like:
int *ptr;
ptr = ptr + 1;
the second line increments the pointer by sizeof(int), which is 2
on MS-DOS machines. Now looking at the mentioned line, it should
be obvious that
(z*(X_DIM * Y_DIM) + y*X_DIM))
calculates the number of array elements This will turn out to be
an arithmetic constant after the calculation. Now since we are
dealing with an array of characters the result of the pointer
arithmetic which adds this value to the pointer to the start of
the array should yield a value equal to the pointer value plus
this constant. Were we using an int data type, i.e. casting our
“space” pointer to (int *), the actual value by which the pointer
would be incremented would be the calculated value times
sizeof(int).
