accessing 2d array using pointers in C

Question

In my main I have

  int mat[ROW][COL];
  int *p = &mat[0][0];
  minput(p, ROW, COL);

I'm stuck on how to make minput, because I'm supposed to initialize the 2d array with set numbers, but I just don't understand how I'm supposed to assign the numbers,

I get that m[0][0] in pointer is *(*(m+0)+0), but it doesn't let me do something like *(*(m+0)+0)=0;,

because it gives the invalid type argument of unary ‘*’ (have ‘int’) error.

So basically I just want to know how to assign a number to a 2d array using pointers, if we didn't have to use pointers this would be so easy with m[2][2]=5 and whatever but I have no idea how to do this with pointers.

My prof provided a one line example of the minput function and it was just *(m+0) = 8;, which doesn't make sense to me, isn't that what we use for a 1d array? if that's the way to do it then does that mean I'm supposed to do something like

*(m+0) = 8;
*(m+1) = 1;
*(m+2) = 6;
*(m+3) = 3;
*(m+4) = 5;
*(m+5) = 7;
*(m+6) = 4;
*(m+7) = 9;
*(m+8) = 2;

for a 3x3 array? it doesn't seem right to me, It would be great if someone could help me out with this, pointers are very confusing to me.

EDIT: so I see that apparently the correct way to do it is the way I listed above with the *(*(m+0)+0) = val

but when I do this in my code, I get the this error:

error: invalid type argument of unary ‘*’ (have ‘int’)
  *(*(m+0)+2)=6;
  ^

why is this the case? is the assignment wrong?

Why not pas a 2D array as a pointer to array? I.e. f(int arr[ROW][COL])? Anything else is just nonsense. (That is converted to int (*)[COL], but you can use 2D indexing as usual). And make p the correct type, too! — too honest for this site
– too honest for this site, Commented Jan 21, 2016 at 2:02
Oh, and if that what your prof provided is really meant for a 2D array, change course. And enable compiler warnings and pay heed to them - especially when it comes to pointer and arrays. — too honest for this site
– too honest for this site, Commented Jan 21, 2016 at 2:05

user3813674 · Accepted Answer · 2016-01-21 00:20:28Z

2

Suppose you have 2D array declared as arr[rows][cols]. Then:

arr points to the first subarray
arr + 1 points to the second subarray
*(*(arr+2)+3) is equivalent to arr[2][3]

So if you want to assign a val to the first element of second row for example:

*(*(mat+1) + 0) = val;

edited Jan 21, 2016 at 0:20

answered Jan 21, 2016 at 0:15

user3813674

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myin528 · Accepted Answer · 2016-01-21 00:12:04Z

1

That's correct. You can use

*(m + (rowNumber*COL) + colNumber) = val

to access it. Basically the elements in the two-dimensional array are stored consecutively in memory, and m points to the beginning of it. Therefore you can do it in this way as an one-dimensional array. That's why we don't use something like **m, as it's not "a pointer to a pointer", even it's a two-dimensional array.

edited Jan 21, 2016 at 0:12

answered Jan 21, 2016 at 0:03

myin528

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3 Comments

anotherdev Over a year ago

you should add that it's why we can't convert 2D Arrays into double pointers (int **), it's a common mistake.

ay lmao Over a year ago

so if I wanted to assign the value 4 to m[2][2], I would do *(m + (2*COL) + 2) = 4 ?

myin528 Over a year ago

That's correct. Though in your case a type cast is needed, which is *(p+...), or alternatively *((int *)mat + ...).

shadowfire · Accepted Answer · 2016-01-21 00:10:36Z

1

The way you have it initialized *((p+0) + 0) = 1; would make mat[0][0] = 1;

*((p+0) + 1) = 2; would be mat[0][1] = 2;
*((p+1) + 0) = 3; would be mat[1][0] = 3;

etc..

answered Jan 21, 2016 at 0:10

shadowfire

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