Thread: How to Realloc 2-D array

Am trying to reallocate a 4x2 matrix to a 6x4 matrix. There is no memory fault but the original values are not retained.

Can someone please tell me if am messing something:

Code:

	int **K1,**K2;
	int i,j,Col=2,Row=4;

	K1 = (int **) calloc(Row,sizeof(int **));

	for (i=1;i<=Row;i++)
		K1[i] = (int *) calloc(Col,sizeof(int));

	for(i=1;i<=Row;i++)
	{
		for(j=1;j<=Col;j++)
		{
			K1[Row][Col] = i+j;
			printf("%d \t",K1[Row][Col]);
		}
		printf("\n");
	}

	if ((K2 = (int **) realloc(K1,Row+2)) == NULL)
			printf("ERROR");
	else
	                       K1 = K2;

	for(i=1;i<=Row;i++)
	{
		if ((K2[i] = (int *) realloc(K1,Col+2)) == NULL)
			printf("\n\n Error...");
		else
			K1[i] = K2[i];
	}

	for(i=Row+1;i<=Row+2;i++)
		K1[i] = (int *) calloc(Col+2,sizeof(int));


	printf("\n");

	for(i=1;i<=Row+2;i++)
	{
		for(j=1;j<=Col+2;j++)
		{
			//K1[Row][Col] = i+j;
			printf("%d \t",K1[Row][Col]);
		}
		printf("\n");
	}

Output:
2 3
3 4
4 5
5 6

0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

You've got serveral errors - including memory faults.

First: Arrays in C start at 0, not 1. In your very first for() loop, i's max value is Row, or 4. You only allocated room for 4 rows however, and K1[i] = K1[Row] = K1[4] accesses the fifth item, which you never allocated memory for. (Since they start at zero, K1[0] = first row, K1[1] = second row, etc.) This happens a lot in your code.

Code:

printf("%d \t",K1[Row][Col]);

Why is that in a loop? Row and Col are constant in your function at that point... so that prints the same value each time. Which means you cannot be getting the output you posted, if I'm reading your code correctly.

Finally, your use of malloc / calloc / realloc. First, they all return void *s, so you do not (and should not) cast them. (Edit: Which dude543 didn't change) Your realloc's error handling also memory leaks - when realloc fails, it doesn't free the old memory. Something like:

Code:

a_type *temp, *actual;
...
temp = realloc(actual, somesize);
if(!temp)
{
  free(actual);
  // continued error handling.
}
actual = temp;

You just need to ensure that actual doesn't become a memory leak - the above example could be modified to not free() actual if it is stored somewhere else, hence the user won't loose their data. All depends on you.

Thnks! have changes the index from 1 to 0, but that still doesnt solve the problem..

I want to retain the 1st values in 4x2 matrix to the 6x4 matrix. Have tried a couple of things with the dummy **K2 but doest help, now the code aborts..

Will appreciate if someone let know.. thx!

Code:

#include<stdio.h>
#include<stdlib.h>


int main(void)
{
int **K1,**K2;
int i,j,Col=2,Row=4;


K1 = (int **) calloc(Row, sizeof(int*)); 

for (i = 0 ; i < Row ;i++)
    K1[i] = (int *) calloc(Col,sizeof(int));

for(i= 0 ; i < Row ;i++)
	for(j = 0 ;j < Col ;j++)
		K1[i][j] = i+j+2;

for(i= 0 ; i < Row ;i++)
	{
	for(j = 0 ;j < Col ;j++)
		printf("%d \t",K1[i][j]);
	printf("\n");
	}


if ((K2 = (int **) realloc(K1,Row+2)) == NULL)
			printf("ERROR");
else
			K1 = K2;


for(i=0;i<Row;i++)
{
	if ((K2[i] = (int *) realloc(K1[i],Col+2)) == NULL)
		printf("\n\n Error...");
	else
		K1[i] = K2[i];
}



for(i=Row;i<Row+2;i++)
	K1[i] = (int *) calloc(Col+2,sizeof(int));


	printf("\n");

	for(i=0;i<Row+2;i++)
	{
		for(j=0;j<Col+2;j++)
		{
			printf("%d \t",K1[i][j]);
		}
		printf("\n");
	}


for(i= 0 ; i < Row+2 ;i++)
{ 
	free(K1[i]);
	free(K2[i]);
}
free(K1);
free(K2);

getchar();
return 0;


}

No. They're freeing the rows one at a time, then once you're done, you free the row holders. That is to say, you allocate the number of rows. Then you allocate each row's contents one chunk at a time. You free in reverse of this. Free each row's contents, then free the row pointers themselves.


Quzah.

Originally Posted by quzah

No. They're freeing the rows one at a time, then once you're done, you free the row holders. That is to say, you allocate the number of rows. Then you allocate each row's contents one chunk at a time. You free in reverse of this. Free each row's contents, then free the row pointers themselves.


Quzah.

Yes and no. Yes, you're right, he does need to free one row at a time, then the row holders. (Like he's doing.) But what I think Maragato was getting at was the segments of code like:

Code:

free(K1);
free(K2);

And

Code:

free(K1[i]);
free(K2[i]);

He only ever uses K2 as a temporary holder for the result of realloc(), and then assigns it to K1 on success. Thus they point at the same data - looking at his code, K2 doesn't point to data that K1 doesn't, and thus doesn't need to be free'd.

wots_guge: Your realloc() calls need fixing: the second parameter is how much memory to allocate, but you're forgeting to multiply by the size of your type. realloc(K1[i],Col+2) should be realloc(K1[i], (Col + 2) * sizeof(int)) for example. (And be sure you check to use the right thing in the sizeof()... sizeof(int) != sizeof(int *) )
(Edit: commenting out the free's mentioned above and changing your reallocs gives me your expected output.)

> it works fine now!
Post it then.
This is a fairly tricky process, and my guess is that you've assigned "works" to "it's just stopped crashing".

Yes, you got it wrong here.
> if ((K2[i] = (int *) realloc (K1[i],(Col+2)*(sizeof(int)))) == NULL)

Since they are the same, your effort to avoid stamping on the old pointer fails.

There is no need to have a "temp" of the same type, you can just use a void* for your temp variable. Plus being void*, there's no chance of you accidentally trying to dereference it without the compiler getting upset.

As in

Code:

void *temp;
temp = realloc( K1[i],(Col+2)*sizeof(int) );
if ( temp != NULL ) {
  K1[i] = temp;
}

And why are you still casting the return result of malloc functions?
http://faq.cprogramming.com/cgi-bin/...&id=1043284351
If, without casts you're getting messages about assigning the result of malloc,
"cannot convert int to pointer" - means you forgot to include stdlib.h, so include it
"cannot convert void* to type*" - means you're using a C++ compiler to compile C, so switch to a C compiler.

> the new allocated memory space using realloc has garbage from heap
Correct - you need to clear the memory yourself.
There is no 'calloc' version of realloc.
In fact, using calloc can be a problem when you're trying to allocate 'initialised' pointers and/or floats (or structures containing them).
http://www.eskimo.com/~scs/c-faq.com/malloc/calloc.html