creating 5X3 array using double astrix

*p is the 0 row
*p+1 is the first row
etc..
and i apply to each row a column

why its not working
??

Code:

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

int main()
{
        int i=0;
    int **p=(int**)malloc(5*sizeof(int*));
        for (i=0;i<5;i++)
        {
          *p+i=(int*)malloc(3*sizeof(int));

        }
        return 0;
}

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Change that to

Code:

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p[i] = malloc ...

---

Don't cast the return value of malloc.

Looks like you forgot the parenthesis (in red).

Quote:

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Originally Posted by transgalactic2

*p is the 0 row
*(p+1) is the first row
etc..
and i apply to each row a column

why its not working
??

Code:

---

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

int main()
{
        int i=0;
    int **p=(int**)malloc(5*sizeof(int*));
        for (i=0;i<5;i++)
        {
          *(p+i)=(int*)malloc(3*sizeof(int));

        }
        return 0;
}

---

---

Quote:

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Originally Posted by itCbitC

Looks like you forgot the parenthesis (in red).

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so this code could be interpretative as p[3][5] and as p[5][3]
??

Quote:

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Originally Posted by transgalactic2

so this code could be interpretative as p[3][5] and as p[5][3]
??

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According to the storage allocations it can be interpreted only as p[5][3].

as i see it
we can say *(p+1) is the first column instead of
*(p+1) is the first row

how the storage allocation contradicts it?

What you intepret as column or row is entirely up to you. However, the dimensions that you allocate is correct for:

Code:

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p[5][3];

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but not for
p[3][5];
[/code]
as you would walk outside the memory allocated here:

Code:

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malloc(3*sizeof(int))

---

if you go beyond 2 in the second index.

--
Mats

*(*(p+2)+1) equals p[2][1]
?

Quote:

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Originally Posted by transgalactic2

*(*(p+2)+1) equals p[2][1]
?

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Reading it from the inside out might help:

p: pointer to pointer to int ie &p[0]
p+2: points 2 elements past p ie &p[2]
*(p+2): dereferences to give p[2]
*(p+2)+1: points to the 2nd element of p[2] ie &p[2][1]
*(*(p+2)+1): dereferences to give p[2][1]

thanks:)

Quote:

---

Originally Posted by transgalactic2

*(*(p+2)+1) equals p[2][1]
?

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for arrays
*(p+i)=p[i]=i[p]=*(i+p)
*(*(p+i)+j)=p[i][j]