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Simple question: Cannot convert 2D array to pointer-to-pointer

P: n/a
I can do this:

int asdf[2];
int* zxcv = asdf;

but not this:

int asdf[2][2];
int** zxcv = asdf;

or I get 'cannot convert from int[2][2] to int**'. Does anybody know why
this is, and how to get around it? I just want a pointer to that double
array. Thanks in advance.
Jul 22 '05 #1
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5 Replies


P: n/a

"overbored" <ov*********@SPAMoverbored.net> wrote in message
news:Xn******************************@127.0.0.1...
I can do this:

int asdf[2];
int* zxcv = asdf;

but not this:

int asdf[2][2];
int** zxcv = asdf;

or I get 'cannot convert from int[2][2] to int**'. Does anybody know why
this is, and how to get around it?
C++ says T[] converts to T*, it doesn't say anything about T[][] to T** and
why should it? It would be impossible to implement. T[] converts to T*
because x[i] is equivalent to *(x + i). But x[i][j] is not equivalent to
*(*(x + i) + j) because the pointer arithmetic doesn't work.
I just want a pointer to that double
array. Thanks in advance.


Why? If you explain what you are trying to do then maybe someone can tell
you how.

You can do this

int asdf[2][2];
int (*zxcv)[2] = asdf;

Is that good enough?

john
Jul 22 '05 #2

P: n/a
I don't understand your explanation. Why do you say the pointer
arithmetic "wouldn't work"? Assuming x : int**,

x
+---+ +---+ +---+---+
| |---->| |---->| 5 | 7 |
+---+ +---+ +---+---+
| |
+---+
int** int* int
So,

x : int** returns the address of the 0th element of the first array.

x + i : int** returns the address of the ith element of the first array.

*(x + i) : int* returns the value of the ith element of the first array,
or the address of the 0th element in the second array.

*(x + i) + j : int* returns the address of the jth element in the second
array.

*(*(x + i) + j) : int returns the value of the jth element in the second
array.

Can you explain what I'm missing? I'm not asserting anything; I'm sure
there's a perfectly good reason why this casting is not allowed. I'm
just not seeing it.

As for my original question, I'm simply wondering how to do the
following:

int** asdf = new int[i][j];

You can't apply the approach you suggested because you don't know the
size of the array in at compile-time.

"John Harrison" <jo*************@hotmail.com> wrote in
news:2q************@uni-berlin.de:

"overbored" <ov*********@SPAMoverbored.net> wrote in message
news:Xn******************************@127.0.0.1...
I can do this:

int asdf[2];
int* zxcv = asdf;

but not this:

int asdf[2][2];
int** zxcv = asdf;

or I get 'cannot convert from int[2][2] to int**'. Does anybody know
why this is, and how to get around it?


C++ says T[] converts to T*, it doesn't say anything about T[][] to
T** and why should it? It would be impossible to implement. T[]
converts to T* because x[i] is equivalent to *(x + i). But x[i][j] is
not equivalent to *(*(x + i) + j) because the pointer arithmetic
doesn't work.
I just want a pointer to that double
array. Thanks in advance.


Why? If you explain what you are trying to do then maybe someone can
tell you how.

You can do this

int asdf[2][2];
int (*zxcv)[2] = asdf;

Is that good enough?

john


Jul 22 '05 #3

P: n/a
Sorry, that was me.

Class Account <cs******@imail.eecs.berkeley.edu> wrote in
news:Xn**********************************@127.0.0. 1:

[snip]
Jul 22 '05 #4

P: n/a
"Class Account" <cs******@imail.eecs.berkeley.edu> wrote in message
news:Xn**********************************@127.0.0. 1...
I don't understand your explanation. Why do you say the pointer
arithmetic "wouldn't work"? Assuming x : int**,

x
+---+ +---+ +---+---+
| |---->| |---->| 5 | 7 |
+---+ +---+ +---+---+
| |
+---+
int** int* int
So,

x : int** returns the address of the 0th element of the first array.

x + i : int** returns the address of the ith element of the first array.

*(x + i) : int* returns the value of the ith element of the first array,
or the address of the 0th element in the second array.

*(x + i) + j : int* returns the address of the jth element in the second
array.

*(*(x + i) + j) : int returns the value of the jth element in the second
array.

Can you explain what I'm missing? I'm not asserting anything; I'm sure
there's a perfectly good reason why this casting is not allowed. I'm
just not seeing it.
A two-dimensional array's elements are stored contiguously, not in separate
arrays that are accessed via pointers that are stored contiguously. When
you access elements using brackets (e.g. [4][5]), you are actually using a
single (computed) offset from the first element. The same syntax for a
pointer-to-pointer, on the other hand, uses one offset to locate a pointer,
and then uses an offset from that pointer to locate the element desired.
They are quite different.
As for my original question, I'm simply wondering how to do the
following:

int** asdf = new int[i][j];

You can't apply the approach you suggested because you don't know the
size of the array in at compile-time.

<snip>

See the FAQ (http://www.parashift.com/c++-faq-lite/), Section 16 ("Freestore
management"), question 15 ("How do I allocate multidimensional arrays using
new?"). In fact, the whole section is a good read if you have questions
about dynamically allocating objects.

--
David Hilsee
Jul 22 '05 #5

P: n/a
> As for my original question, I'm simply wondering how to do the
following:

int** asdf = new int[i][j];


First things first:

int* a = new int[125];
is lain out in memory EXACTLY as is:
int (*b)[5][5] = new int[5][5][5];
Now, for some reason you want a pointer to a pointer to that array.

int** p_p_ints;

Okay, so in this variable, we need to store that address of a pointer. But
you don't have a pointer whose address to put in! Make one:

int** p_p_ints; //Pointer to pointer to int

int* p_ints; //pointer to int

p_ints = new int[5][5][5];

p_p_ints = &p_ints;
Now p_p_ints points to a pointer which points to your array of ints. And why
exactly do you want this?!
Once you've defined an array like:

int monkey[60];
there are ways of accessing it as if it were:
int monkey[2][30]

int monkey[3][20]

int monkey[4][15]

int monkey[5][12]

int monkey[6][10]
-JKop
Jul 22 '05 #6

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