Re: [algogeeks] Re: 2 D array(dynamic allocation)
Hi Gene,
I was thinking the same thing which you implemented in your first
snippet. But I tried it without using typedef. Following is my code:
#include stdio.h
#include malloc.h
int main()
{
int (*ptr)[4] ;
ptr = (int ((*)[4]))malloc(2*sizeof(int (*)[4]));
if (ptr) {
int i,j,cnt =0;
for (i= 0; i 2;i++)
for (j = 0; j 4; j++)
ptr[i][j] = cnt++;
for (i= 0; i 2;i++)
for (j = 0; j 4; j++)
printf(%d\n,ptr[i][j]);
/* Free will not work, as the above code
* overwrites the signature used by free
*/
/* free((ptr)); */
}
return 0;
}
This code is buggy, because as soon as the malloc returns, I try to
write some data and in that process, I am overwriting the signature
(which is used by free()), and hence free() call fails. Can you point
out the problem?
Thanks,
Vishal
On Sun, Jul 3, 2011 at 2:12 AM, Gene gene.ress...@gmail.com wrote:
Unfortunately this invokes undefined behavior, so it may not work for
all architectures and compilers. It relies on pointers to int having
the same alignment as int.
By far the best way to do this is use C99 features:
double (*a)[n_cols] = calloc(n_rows, sizeof *a);
If you don't have C99 and the row length is fixed, then you can
allocated a 2d array very easily:
#include stdio.h
#include stdlib.h
#define ROW_SIZE 10
typedef double ROW[ROW_SIZE];
int main(void)
{
int i, j n_rows = 15;
ROW *array = malloc(n_rows * sizeof(ROW));
for (i = 0; i n_rows; i++)
for (j = 0; j ROW_SIZE; j++)
array[i][j] = 100 * i + j;
for (i = 0; i n_rows; i++) {
for (j = 0; j ROW_SIZE; j++)
printf(%5.0f, array[i][j]);
printf(\n);
}
return 0;
}
If you need both array axes to be variable, then the norm is to
allocate a 1d array and define a macro to allow 2d access:
#include stdio.h
#include stdlib.h
typedef double *ARRAY;
#define Elt(A, I, J, NCOLS) ((A)[(I) * (NCOLS) + (J)])
int no_fixed(void)
{
int i, j, n_rows = 15, n_cols = 10;
ARRAY array = malloc(n_rows * n_cols * sizeof array[0]);
for (i = 0; i n_rows; i++)
for (j = 0; j n_cols; j++)
Elt(array, i, j, n_cols) = 100 * i + j;
for (i = 0; i n_rows; i++) {
for (j = 0; j n_cols; j++)
printf(%5.0f, Elt(array, i, j, n_cols));
printf(\n);
}
return 0;
}
On Jul 2, 2:05 pm, vaibhav shukla vaibhav200...@gmail.com wrote:
@sandeep sir: thnx... good 1 :)
On Sat, Jul 2, 2011 at 11:32 PM, Sandeep Jain sandeep6...@gmail.com wrote:
Here's my solution.
int** allocateMatrix(int m, int n)
{
int **rowList = (int**)malloc(sizeof(int)*m*n + sizeof(int*)*m);
int *colList = (int*)(rowList+m);
int i;
for(i=0; im; i++)
{
rowList[i] = colList+i*n;
}
return rowList;
}
And here's the main method to test/understand the allocation.
int main()
{
int m=3, n=4;
int **mat = allocateMatrix(m,n);
int i, j, c=0;
int wordCount= m*n+m; //sizeof(int)*4*5 + sizeof(int*)*4; counting
words so sizeof is not needed
int* memMap = (int*)mat;
// Fill array elements with some values to be able to test
for(i=0; im; i++)
for(j=0; jn; j++)
mat[i][j] = c++;
printf(\nAddress\t Value\n);
for(i=0; iwordCount; i++)
printf(\n%u\t== %u, memMap+i, memMap[i]);
getchar();
}
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[algogeeks] Re: 2 D array(dynamic allocation)
Here's my solution.
int** allocateMatrix(int m, int n)
{
int **rowList = (int**)malloc(sizeof(int)*m*n + sizeof(int*)*m);
int *colList = (int*)(rowList+m);
int i;
for(i=0; im; i++)
{
rowList[i] = colList+i*n;
}
return rowList;
}
And here's the main method to test/understand the allocation.
int main()
{
int m=3, n=4;
int **mat = allocateMatrix(m,n);
int i, j, c=0;
int wordCount= m*n+m; //sizeof(int)*4*5 + sizeof(int*)*4; counting
words so sizeof is not needed
int* memMap = (int*)mat;
// Fill array elements with some values to be able to test
for(i=0; im; i++)
for(j=0; jn; j++)
mat[i][j] = c++;
printf(\nAddress\tValue\n);
for(i=0; iwordCount; i++)
printf(\n%u\t== %u, memMap+i, memMap[i]);
getchar();
}
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Re: [algogeeks] Re: 2 D array(dynamic allocation)
@sandeep sir: thnx... good 1 :)
On Sat, Jul 2, 2011 at 11:32 PM, Sandeep Jain sandeep6...@gmail.com wrote:
Here's my solution.
int** allocateMatrix(int m, int n)
{
int **rowList = (int**)malloc(sizeof(int)*m*n + sizeof(int*)*m);
int *colList = (int*)(rowList+m);
int i;
for(i=0; im; i++)
{
rowList[i] = colList+i*n;
}
return rowList;
}
And here's the main method to test/understand the allocation.
int main()
{
int m=3, n=4;
int **mat = allocateMatrix(m,n);
int i, j, c=0;
int wordCount= m*n+m; //sizeof(int)*4*5 + sizeof(int*)*4; counting
words so sizeof is not needed
int* memMap = (int*)mat;
// Fill array elements with some values to be able to test
for(i=0; im; i++)
for(j=0; jn; j++)
mat[i][j] = c++;
printf(\nAddress\tValue\n);
for(i=0; iwordCount; i++)
printf(\n%u\t== %u, memMap+i, memMap[i]);
getchar();
}
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best wishes!!
Vaibhav Shukla
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[algogeeks] Re: 2 D array(dynamic allocation)
Unfortunately this invokes undefined behavior, so it may not work for
all architectures and compilers. It relies on pointers to int having
the same alignment as int.
By far the best way to do this is use C99 features:
double (*a)[n_cols] = calloc(n_rows, sizeof *a);
If you don't have C99 and the row length is fixed, then you can
allocated a 2d array very easily:
#include stdio.h
#include stdlib.h
#define ROW_SIZE 10
typedef double ROW[ROW_SIZE];
int main(void)
{
int i, j n_rows = 15;
ROW *array = malloc(n_rows * sizeof(ROW));
for (i = 0; i n_rows; i++)
for (j = 0; j ROW_SIZE; j++)
array[i][j] = 100 * i + j;
for (i = 0; i n_rows; i++) {
for (j = 0; j ROW_SIZE; j++)
printf(%5.0f, array[i][j]);
printf(\n);
}
return 0;
}
If you need both array axes to be variable, then the norm is to
allocate a 1d array and define a macro to allow 2d access:
#include stdio.h
#include stdlib.h
typedef double *ARRAY;
#define Elt(A, I, J, NCOLS) ((A)[(I) * (NCOLS) + (J)])
int no_fixed(void)
{
int i, j, n_rows = 15, n_cols = 10;
ARRAY array = malloc(n_rows * n_cols * sizeof array[0]);
for (i = 0; i n_rows; i++)
for (j = 0; j n_cols; j++)
Elt(array, i, j, n_cols) = 100 * i + j;
for (i = 0; i n_rows; i++) {
for (j = 0; j n_cols; j++)
printf(%5.0f, Elt(array, i, j, n_cols));
printf(\n);
}
return 0;
}
On Jul 2, 2:05 pm, vaibhav shukla vaibhav200...@gmail.com wrote:
@sandeep sir: thnx... good 1 :)
On Sat, Jul 2, 2011 at 11:32 PM, Sandeep Jain sandeep6...@gmail.com wrote:
Here's my solution.
int** allocateMatrix(int m, int n)
{
int **rowList = (int**)malloc(sizeof(int)*m*n + sizeof(int*)*m);
int *colList = (int*)(rowList+m);
int i;
for(i=0; im; i++)
{
rowList[i] = colList+i*n;
}
return rowList;
}
And here's the main method to test/understand the allocation.
int main()
{
int m=3, n=4;
int **mat = allocateMatrix(m,n);
int i, j, c=0;
int wordCount= m*n+m; //sizeof(int)*4*5 + sizeof(int*)*4; counting
words so sizeof is not needed
int* memMap = (int*)mat;
// Fill array elements with some values to be able to test
for(i=0; im; i++)
for(j=0; jn; j++)
mat[i][j] = c++;
printf(\nAddress\t Value\n);
for(i=0; iwordCount; i++)
printf(\n%u\t== %u, memMap+i, memMap[i]);
getchar();
}
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Re: [algogeeks] Re: 2 D array(dynamic allocation)
do we need to add m in each a[i]?? shudnt it be a[i] = a+ n*i?? i think i m
doing a silly mistake.. bt plz tell me..
On Thu, Jun 30, 2011 at 7:41 AM, Dave dave_and_da...@juno.com wrote:
@Rizwan: Not completely. What if ROW in your code is a variable (not a
constant) that is not known until run time? Then you need to
dynamically allocate space for your arr2D as well. So, contradicting
my earlier No response, maybe something like this would work to
allocate an array a with m rows and n columns:
int* a = (int*)malloc( m * ( sizeof(int*) + n * sizeof(int) ) );
for( i = 0 ; i m ; ++i )
a[i] = a + m + i * n;
Dave
On Jun 29, 4:41 pm, rizwan hudda rizwanhu...@gmail.com wrote:
I have solved this using one malloc find the code inhttp://
ideone.com/BV9Kj
On Thu, Jun 30, 2011 at 1:20 AM, Piyush Sinha ecstasy.piy...@gmail.com
wrote:
ohh sorrymy bad...i didnt read the whole question..i just read the
subject...:P
i think its not possible if u want other than hary's solution...
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
@piyush: only one call to malloc...ur sol has 2
On Thu, Jun 30, 2011 at 12:58 AM, Piyush Sinha
ecstasy.piy...@gmail.comwrote:
int **p;
p = (int **)malloc(sizeof(int *)*row);
for(i = 0;irow;i++)
p[i] = (int *)malloc(sizeof(int)*column);
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
though thankx :)
On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
apoorvemo...@gmail.comwrote:
@above: man i need a 2d array not a 1d array...
On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
harry.rat...@gmail.comwrote:
#includestdlib.h
int main ()
{
int *mat;
int i,j;
int ROW=4;
int COL=3;
int k=0;
mat=(int *)malloc(ROW*COL*sizeof(int));
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
mat[i*COL+j]=++k;
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
printf(%d,,mat[i*COL+j]);
return 0;
}
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Re: [algogeeks] Re: 2 D array(dynamic allocation)
#includestdio.h
int main()
{
int i,j;
int ct=0;
int *a[3];
for( i=0;i3;i++)
{
a[i]=(int **)malloc(sizeof(int)*4);
for(i=0;i3;i++){
for(j=0;j4;j++)
a[i][j]=ct++;
printf(%d,a[i][j]);
}
}
}
On Sat, Jul 2, 2011 at 12:13 AM, Anika Jain anika.jai...@gmail.com wrote:
do we need to add m in each a[i]?? shudnt it be a[i] = a+ n*i?? i think i m
doing a silly mistake.. bt plz tell me..
On Thu, Jun 30, 2011 at 7:41 AM, Dave dave_and_da...@juno.com wrote:
@Rizwan: Not completely. What if ROW in your code is a variable (not a
constant) that is not known until run time? Then you need to
dynamically allocate space for your arr2D as well. So, contradicting
my earlier No response, maybe something like this would work to
allocate an array a with m rows and n columns:
int* a = (int*)malloc( m * ( sizeof(int*) + n * sizeof(int) ) );
for( i = 0 ; i m ; ++i )
a[i] = a + m + i * n;
Dave
On Jun 29, 4:41 pm, rizwan hudda rizwanhu...@gmail.com wrote:
I have solved this using one malloc find the code inhttp://
ideone.com/BV9Kj
On Thu, Jun 30, 2011 at 1:20 AM, Piyush Sinha ecstasy.piy...@gmail.com
wrote:
ohh sorrymy bad...i didnt read the whole question..i just read the
subject...:P
i think its not possible if u want other than hary's solution...
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
@piyush: only one call to malloc...ur sol has 2
On Thu, Jun 30, 2011 at 12:58 AM, Piyush Sinha
ecstasy.piy...@gmail.comwrote:
int **p;
p = (int **)malloc(sizeof(int *)*row);
for(i = 0;irow;i++)
p[i] = (int *)malloc(sizeof(int)*column);
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
though thankx :)
On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
apoorvemo...@gmail.comwrote:
@above: man i need a 2d array not a 1d array...
On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
harry.rat...@gmail.comwrote:
#includestdlib.h
int main ()
{
int *mat;
int i,j;
int ROW=4;
int COL=3;
int k=0;
mat=(int *)malloc(ROW*COL*sizeof(int));
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
mat[i*COL+j]=++k;
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
printf(%d,,mat[i*COL+j]);
return 0;
}
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regards
Apoorve Mohan
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[algogeeks] Re: 2 D array(dynamic allocation)
@saurabh.. No man.. it involves variable number of calls.. even if its present only once in code. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/2VonkRB6WMEJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: 2 D array(dynamic allocation)
#includestdio.h
int main()
{
int i,j;
int ct=0;
int *a[3];
for( i=0;i3;i++)
{
a[i]=(int **)malloc(sizeof(int)*4);
}
for(i=0;i3;i++){
for(j=0;j4;j++)
a[i][j]=ct++;
printf(%d,a[i][j]);
}
On Sat, Jul 2, 2011 at 2:27 AM, sravanreddy001 sravanreddy...@gmail.comwrote:
@saurabh.. No man.. it involves variable number of calls.. even if its
present only once in code.
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[algogeeks] Re: 2 D array(dynamic allocation)
Apoorve: No.
Dave
On Jun 29, 2:34 pm, Apoorve Mohan apoorvemo...@gmail.com wrote:
@piyush: only one call to malloc...ur sol has 2
On Thu, Jun 30, 2011 at 12:58 AM, Piyush Sinha
ecstasy.piy...@gmail.comwrote:
int **p;
p = (int **)malloc(sizeof(int *)*row);
for(i = 0;irow;i++)
p[i] = (int *)malloc(sizeof(int)*column);
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
though thankx :)
On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
apoorvemo...@gmail.comwrote:
@above: man i need a 2d array not a 1d array...
On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
harry.rat...@gmail.comwrote:
#includestdlib.h
int main ()
{
int *mat;
int i,j;
int ROW=4;
int COL=3;
int k=0;
mat=(int *)malloc(ROW*COL*sizeof(int));
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
mat[i*COL+j]=++k;
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
printf(%d,,mat[i*COL+j]);
return 0;
}
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regards
Apoorve Mohan
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regards
Apoorve Mohan
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regards
Apoorve Mohan- Hide quoted text -
- Show quoted text -
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[algogeeks] Re: 2 D array(dynamic allocation)
@Rizwan: Not completely. What if ROW in your code is a variable (not a
constant) that is not known until run time? Then you need to
dynamically allocate space for your arr2D as well. So, contradicting
my earlier No response, maybe something like this would work to
allocate an array a with m rows and n columns:
int* a = (int*)malloc( m * ( sizeof(int*) + n * sizeof(int) ) );
for( i = 0 ; i m ; ++i )
a[i] = a + m + i * n;
Dave
On Jun 29, 4:41 pm, rizwan hudda rizwanhu...@gmail.com wrote:
I have solved this using one malloc find the code inhttp://ideone.com/BV9Kj
On Thu, Jun 30, 2011 at 1:20 AM, Piyush Sinha ecstasy.piy...@gmail.com
wrote:
ohh sorrymy bad...i didnt read the whole question..i just read the
subject...:P
i think its not possible if u want other than hary's solution...
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
@piyush: only one call to malloc...ur sol has 2
On Thu, Jun 30, 2011 at 12:58 AM, Piyush Sinha
ecstasy.piy...@gmail.comwrote:
int **p;
p = (int **)malloc(sizeof(int *)*row);
for(i = 0;irow;i++)
p[i] = (int *)malloc(sizeof(int)*column);
On 6/30/11, Apoorve Mohan apoorvemo...@gmail.com wrote:
though thankx :)
On Thu, Jun 30, 2011 at 12:44 AM, Apoorve Mohan
apoorvemo...@gmail.comwrote:
@above: man i need a 2d array not a 1d array...
On Thu, Jun 30, 2011 at 12:38 AM, hary rathor
harry.rat...@gmail.comwrote:
#includestdlib.h
int main ()
{
int *mat;
int i,j;
int ROW=4;
int COL=3;
int k=0;
mat=(int *)malloc(ROW*COL*sizeof(int));
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
mat[i*COL+j]=++k;
for(i=0;iROW;i++)
for(j=0;jCOL;j++)
printf(%d,,mat[i*COL+j]);
return 0;
}
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regards
Apoorve Mohan
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regards
Apoorve Mohan
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regards
Apoorve Mohan
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Thanks and regards
Rizwan A Huddahttp://sites.google.com/site/rizwanhudda2- Hide quoted text -
- Show quoted text -
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