two dimensional Arrays access using pointers

Question

Is

 *(ary[3]+8)

and

 ary[3][8]

are the same ? If yes, please explain how ? ary[3] returns the address of first element or the value in ary[3][0] ? ary is a two dimensional array.

Thanks in advance.

You can access also it like this: `*(*(ary+ 3) + 8)`. Note that `ary[3]` is the same as `*(ary + 3)`. This should answer your question. — Maroun, Jul 02 '13 at 07:08
The value at *(ary+3) and ary[3] are the same. What does *(ary+3) return ? The value at ary[3][0] or the starting address of ary[3] ? — Srini Vas, Jul 02 '13 at 07:10
@StoryTeller : ary is 2d array. Can you please share your knowledge on this ? — Srini Vas, Jul 02 '13 at 07:16
@SriniVas, see here http://stackoverflow.com/a/14111286/817643 about what defining a true 2d array means — StoryTeller - Unslander Monica, Jul 02 '13 at 09:03

score 4 · Accepted Answer · edited Jul 02 '13 at 07:09

4

Yes

a[i] is same as *(a+i)

ary[i][j] is same as *( *(ary+i)+j))

edited Jul 02 '13 at 07:09

Maroun

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answered Jul 02 '13 at 07:09

banarun

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Doesn't *(ary+i) in *(*(ary+i)+j)) return value at ary[i] ? Pointer holds an address and * is to retrieve the value at the address. Going by this, *(ary+i) has to return the value in ary+i. And we add j to the value at the address ary+i and change the value at the address obtained from the previous step. Isn't that how it has to work or works ? – Srini Vas Jul 02 '13 at 07:15
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I do not see a contradiction in the previous Srini Vas's comment. He says pointer holds and address and * is to retrive the value at the address. The value, in this case, is a pointer to a (1dim) array. – ondrejdee Jul 02 '13 at 07:24
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@StoryTeller: What is a counterexample? – Oliver Charlesworth Jul 02 '13 at 07:26
A true 2d array `a[2][3]`, when indexed, expands into `*(a + 2i + j)`. The size is part of the type. What banarun said is only true when `a` is a one dimensional array of pointers. – StoryTeller - Unslander Monica Jul 02 '13 at 08:58
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@StoryTeller: It's *always* the case that `a[i][j]` is equivalent to `*(*(a+i) + j)`. For a true 2D array, this is *also* equivalent to `*((T*)a + 3*i + j)` (where `T` is the type of each element). – Oliver Charlesworth Jul 02 '13 at 14:19
@OliCharlesworth, if that's the case, `*(a+i)` must be a pointer type. What type is it? – StoryTeller - Unslander Monica Jul 02 '13 at 14:53
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@StoryTeller: If you have `T a[M][N]`, then `*(a+i)` is of type `T[N]`, which decays to `T*` in most situations. – Oliver Charlesworth Jul 02 '13 at 15:55
@OliCharlesworth, ignore the deleted comment. I removed it for a reason :) – StoryTeller - Unslander Monica Jul 02 '13 at 16:04

Dayal rai · Answer 2 · 2013-07-02T07:44:11.553

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*(ary[3]+8) says value at 8th column of third row.ary[3] is base address of third Row.ary[3][8] will also access to same element at third row and 8th column.

For Example i am taking an 2D array of two row and 4 column which is equivalent to 1D array of 8 elements.As shown below.

int a[8] = {0,1,2,3,4,5,6,7};

int b[2][4] = {{0,1,2,3},{4,5,6,7}};

since b is 2D array , so you can consider it as array of two 1D arrays.when you pass b[1] or b[1][0] it says address of first row.Rectangular array allocated in memory by Row.so if you want to find address of element a[row][col] it will get calculated as

address = baseAddress + elementSize * (row*(total number of column) + col);

edited Jul 02 '13 at 07:44

answered Jul 02 '13 at 07:14

Dayal rai

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Where/How in memory is the base array of ary[3] is stored? ary[0..n][0..n] are stored sequentially. ary[3] and ary[3][0] has to have the same base address and how is it resolved when invoked by calls of above types? – Srini Vas Jul 02 '13 at 07:22
@SriniVas Edited my answer for clarity. – Dayal rai Jul 02 '13 at 07:39
@SriniVas Aha, this is what you mean. The values of `ary[i]` are not stored anywhere if your array is declared as e.g. `int ary[4][9]`. The compiler knows that your `ary` is a two-dimensional array, and when you do `ary[i]` this gives the address you say. But note that the following example is completely different case: `int *ary[4]; int ary1[9]; int ary2[9]; int ary3[9]; int ary4[9]; ary[0] = ary1; ary[1] = ary2; ary[2] = ary3; ary[3] = ary4;` although you can indeed call this a 2d array as well. Here, the pointers to the second dimension are explicitly stored. – ondrejdee Jul 02 '13 at 07:48

score 1 · Answer 3 · answered Jul 02 '13 at 07:17

If x is an array (int, say) x[i] is just a syntactic sugar for *(x+i). In your case, ary is a two-dimensional array (again of int, say). By the same syntactic sugar mechanism, ary[i][j] is equivalent to *((*(ary+i))+j), from which it is clear what happens under the hood.

score 0 · Answer 4 · answered Jul 02 '13 at 08:46

As others already have said, a[i] is just a sugar for *(a+i).

I just would like to add that it always works, that allows us to do things like that:

char a[10];
char b;
char c[10][20];

// all of these are the same:
b = a[5];      // classic
b = *(a + 5);  // pointer shifting
b = 5[a];      // even so!

b = c[5][9];
b = *(c[5] + 9);
b = *(*(c + 5) + 9);
b = *(c + 5)[9];
b = 5[c][9];
b = 5[9][c];  // WRONG! Compiling error

two dimensional Arrays access using pointers

4 Answers4