2D array to pointers

Question

I've got array: int i[20][30]. What is the fastest way to make int** pointer to use on it?

My friend just wrote function which uses same size arrays and my function is on pointers and I need to pass his data.

The fastest way is don't. Anyway why not just use `std::array`? — , Mar 12 '13 at 13:37
A 2d array simply doesn't have the same memory layout as a pointer to pointer would expect. — Joseph Mansfield, Mar 12 '13 at 13:42

Paul R · Answer 1 · 2013-03-12T13:46:04.353

2

int a[20][30];       // data as 2D array
int *ap[20];         // array of pointers
for (j = 0; j < 20; ++j)
    ap[j] = a[j];    // convert array to pointers
// can now pass `ap` to function expecting `int **`...

edited Mar 12 '13 at 13:46

answered Mar 12 '13 at 13:39

Paul R

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@Daniel: thanks - the use of `i` for an array name muddled my brain - now fixed. – Paul R Mar 12 '13 at 13:41
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Why would one downvote this answer? Has anybody a better idea to make a function expecting an `int**` with an `int[20][30]`? – Daniel Fischer Mar 12 '13 at 13:42
Ain't there a faster way? – oneat Mar 12 '13 at 13:43
@oneat: "faster" ? what do you mean by "faster" ? – Paul R Mar 12 '13 at 13:43
I believe that `ip` should be `int *ip[30];`, no? – Joseph Mansfield Mar 12 '13 at 13:43
@sftrabbit No, `int i[20][30]` is an array of 20 `int[30]`, so 20 is correct. – Daniel Fischer Mar 12 '13 at 13:44
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@oneat You're asking to convert between two incompatible things. That kind of conversion takes some effort. – Joseph Mansfield Mar 12 '13 at 13:45
@sftrabbit: no, it's 20 rows of 30 columns (or 20 strings of 30 characters if you prefer). – Paul R Mar 12 '13 at 13:45
@DanielFischer Oh, of course, I'm also getting mixed up with these variable names. – Joseph Mansfield Mar 12 '13 at 13:45
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I've changed the array/pointer names above now as it was somewhat confusing. – Paul R Mar 12 '13 at 13:46

teppic · Answer 2 · 2013-03-12T14:43:04.437

You can't have an int ** pointer to that array directly, it wouldn't make any sense. If you want to access the array as a block of memory you can just use:

int *p = (int *)i; // you can go to up (p + 20*30)

Otherwise you need a pointer to an array of int[30], i.e.

int (*p)[30] = &i[0]; // points to first row

With the first one, because of the cast, you can then define:

int **p2 = &p;

And you can access the array with *(*p2 + n). You've cast away the array dimensions, but as long as you use them in the pointer calculations, it's all valid, i.e. *(*p2 + 35) refers to row 2, column 6 (one row of 30, plus element 5 of the next). This is very quick, but if the function needs to operate on the array using the original dimensions, this wouldn't work (since it isn't an array of pointers).

score 0 · Answer 3 · answered Mar 12 '13 at 13:57

you cant change int i[20][30] to int ** except of manually go over the data and build it.
but, something that may help you, every xD array in c++ is just a 1D array with a somehow complexed way to access a member.

Example:

#define MaxX 20
#define MaxY 30
int i[MaxX][MaxY];
...
//When you write:
i[x][y] = something;
//It's actually been translate to:
i[x * MaxY + y] = something;

2D array to pointers

3 Answers3