C++ 2d array created by double pointer

Question

I want to create 2d array using pointer and show then show it. I change rows by increment pointer (one *) and columns by index. My code:

int ** tabl=new int *[4];
for (int i=0;i<10;i++)
    tabl[i]=new int [3];
int **ptab=tabl;//save first position of array

for (int i=0;i<4;i++)
{
    for (int j=0;j<3;j++)
    {
        *tabl[j]=rand()%10 +1;
        printf("%4d",*tabl[j]);
    }
    tabl++;
}
tabl=ptab;
cout<<endl<<endl<<endl;
for (int i=0;i<4;i++)
{
    for (int j=0;j<3;j++)
    {
        printf("%4d",*tabl[j]);
    }
    tabl++;
}

Output:

   2   8   5   1  10   5   9   9   3   5   6   6

   2   1   9   1   9   5   9   5   6   5   6   6

But some of the digits in the second loop are different then the original. Why?

*tabl[j] means tabl[j][0]. You actually wanted tabl[0][j]. It would be better to use tabl[i][j] instead, and not do tabl++, and not have ptab. — M.M
– M.M, Commented Dec 3, 2015 at 21:54
I know two index method, but my teacher from studies demand using pointers. — diego9403
– diego9403, Commented Dec 3, 2015 at 21:57
This does use pointers. When you write tabl[i], actually tabl decays to a pointer, and the code is defined as being equivalent to *(tabl+i). — M.M
– M.M, Commented Dec 3, 2015 at 22:00

Ishamael · Accepted Answer · 2015-12-03 21:50:51Z

5

The problem is the way you refer to the element. It should be (*tabl)[j], not *tabl[j].

Also note that you go beyound allocated memory here:

int ** tabl=new int *[4];
for (int i=0;i<10;i++)
               ^
    -----------+

answered Dec 3, 2015 at 21:50

Ishamael

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2 Comments

diego9403 Over a year ago

I changed size from 10 to 4, because I really didn't see all values, that's why "10" left(I missed it). Of course I corrected it. I understand the brackets are there because of the priorities?

Ishamael Over a year ago

Yes, *tabl[j] means "dereference j-th element of tabl", not "j-th element of dereferenced table"

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