DaniWeb Forum Index > Software Development > C++

C++ RSS

Dynamic Mmeory Allocation with 2D Arrays

Please support our C++ advertiser: Intel Parallel Studio Home

•

Join Date: Mar 2007

Posts: 53

Reputation: FoX_ is an unknown quantity at this point

Solved Threads: 7

FoX_

Offline

Junior Poster in Training

Dynamic Mmeory Allocation with 2D Arrays

Apr 14th, 2007

Hi all;

I've to use a dynamic array(2D) in my program in C.

I've read some tutorials about this and I understood its mental.But still there exist a few points which I didn't understand.

Here is a code:

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
int main(){
int **a, x;
 
a = (int **)malloc(sizeof(int) * 10);
for(x = 0; x < 10; x ++) {
    a[x] = (int *)malloc(sizeof(int) * 3);
}
 
/* ... */
 
for(x = 0; x < 10; x ++) {
    free(a[x]);
}
free(a);
int main(){
int **a, x;

a = (int **)malloc(sizeof(int) * 10);
for(x = 0; x < 10; x ++) {
    a[x] = (int *)malloc(sizeof(int) * 3);
}

/* ... */

for(x = 0; x < 10; x ++) {
    free(a[x]);
}
free(a);

Firstly: Does the **a mean that first * points the row and second * points the column???(If not what does it mean???)

Secondly: In the "a[x] = (int *)malloc(sizeof(int) * 3);" , malloc function returns an adress and as far as I can see it assigns a[x] to a[x] 's adress.
But when I print the a[x] value and it's adress; I saw it doesn't process like my idea which was indicated above.So what is the problem???

Thanks for your helps...

Last edited by FoX_; Apr 14th, 2007 at 8:10 pm.

•

Join Date: Apr 2006

Posts: 5,051

Reputation: John A is a splendid one to behold

Solved Threads: 332

John A

Offline

Vampirical Lurker

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 14th, 2007

>Does the **a mean that first * points the row and second * points the column???
Almost. The first * points to a row, however the second * points to an individual "cell" if you will, of a row. It points to the integer.

>malloc function returns an adress and as far as I can see it assigns a[x] to a[x] 's adress.
It assigns the address of the freshly-allocated memory to a[x], yes.

>But when I print the a[x] value and it's adress; I saw it doesn't process like my idea which was indicated above.
What do you mean? a[x] just holds a pointer to the actual memory, you'll have to dereference the address contained inside a[x] if you want to print out the value.

"Technological progress is like an axe in the hands of a pathological criminal."

•

Join Date: Mar 2007

Posts: 53

Reputation: FoX_ is an unknown quantity at this point

Solved Threads: 7

FoX_

Offline

Junior Poster in Training

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 14th, 2007

I accept I babbled a little but when I thought the 2D arrays' and pointers' mental I really understood it.
Thanks for your helps...

•

Join Date: Jun 2006

Posts: 7,611

Reputation: ~s.o.s~ has much to be proud of

Solved Threads: 465

~s.o.s~

Offline

Failure as a human

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

•

Originally Posted by FoX_

Help with Code Tags

(Toggle Plain Text)
a = (int **)malloc(sizeof(int) * 10);
for(x = 0; x < 10; x ++) {
    a[x] = (int *)malloc(sizeof(int) * 3);
}
a = (int **)malloc(sizeof(int) * 10); for(x = 0; x < 10; x ++) { a[x] = (int *)malloc(sizeof(int) * 3); }

In this case you were lucky enough to have the size of integer the same as the size of a pointer. If it would have been array of doubles, you would have messed big time.

a = malloc(sizeof(int*) * 10);

A better way would have been:

 Help with Code Tags 
 c Syntax (Toggle Plain Text) 
int main()
{
    const int SIZE = 10;
 
    double **arr = (double**)malloc(SIZE * sizeof(*arr));
    for(int i = 0; i < SIZE; ++i)
    {
        arr[i] = (double*)malloc(SIZE * sizeof(*arr[0]));
    }
 
    for(int i = 0; i < SIZE; ++i)
        for(int j = 0; j < SIZE; ++j)
            arr[i][j] = i * j + i + j;
 
    for(int i = 0; i < SIZE; ++i)
    {
        for(int j = 0; j < SIZE; ++j)
            cout << arr[i][j] << "\t";
        cout << '\n';
    }
    getchar();
    return 0;
}
int main()
{
    const int SIZE = 10;

    double **arr = (double**)malloc(SIZE * sizeof(*arr));
    for(int i = 0; i < SIZE; ++i)
    {
        arr[i] = (double*)malloc(SIZE * sizeof(*arr[0]));
    }

    for(int i = 0; i < SIZE; ++i)
        for(int j = 0; j < SIZE; ++j)
            arr[i][j] = i * j + i + j;

    for(int i = 0; i < SIZE; ++i)
    {
        for(int j = 0; j < SIZE; ++j)
            cout << arr[i][j] << "\t";
        cout << '\n';
    }
    getchar();
    return 0;
}

I don't accept change; I don't deserve to live.

•

Join Date: Apr 2006

Posts: 5,051

Reputation: John A is a splendid one to behold

Solved Threads: 332

John A

Offline

Vampirical Lurker

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

Heh, I completely missed that little detail...

"Technological progress is like an axe in the hands of a pathological criminal."

•

Join Date: Jun 2006

Posts: 7,611

Reputation: ~s.o.s~ has much to be proud of

Solved Threads: 465

~s.o.s~

Offline

Failure as a human

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

Just to let you know that the old smileys are deprecated, you are better off using the new ones, the ones which recent compiler..err Daniweb currently supports... ;-)

I don't accept change; I don't deserve to live.

•

Join Date: Mar 2007

Posts: 120

Reputation: ft3ssgeek is an unknown quantity at this point

Solved Threads: 7

ft3ssgeek

Offline

Junior Poster

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

•

Originally Posted by ~s.o.s~

Just to let you know that the old smileys are deprecated, you are better off using the new ones, the ones which recent compiler..err Daniweb currently supports... ;-)

So, uh, in other words, don't follow your lead??

•

Join Date: Jun 2006

Posts: 7,611

Reputation: ~s.o.s~ has much to be proud of

Solved Threads: 465

~s.o.s~

Offline

Failure as a human

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

Yes, you got that right my friend..

I don't accept change; I don't deserve to live.

•

Join Date: Apr 2006

Posts: 5,051

Reputation: John A is a splendid one to behold

Solved Threads: 332

John A

Offline

Vampirical Lurker

Re: Dynamic Mmeory Allocation with 2D Arrays

Apr 15th, 2007

•

Originally Posted by ~s.o.s~

Just to let you know that the old smileys are deprecated, you are better off using the new ones, the ones which recent compiler..err Daniweb currently supports... ;-)

Nooooo.....

"Technological progress is like an axe in the hands of a pathological criminal."

•

Join Date: Dec 2006

Posts: 1,089

Reputation: vijayan121 is a name known to all

Solved Threads: 164

vijayan121 vijayan121 is offline

Offline

Veteran Poster

Re: Dynamic Mmeory Allocation with 2D Arrays

#10

Apr 15th, 2007

•

Originally Posted by ~s.o.s~

const int SIZE = 10;
double **arr = (double**)malloc(SIZE * sizeof(*arr));
for(int i = 0; i < SIZE; ++i)
{
arr[i] = (double*)malloc(SIZE * sizeof(*arr[0]));
}
}[/code]

This is first allocating an array of pointers, and allocating memory one by one for each one dimensional
array. (this kind of array is called a jagged array in C#).
A true two dimensional array is rectangular; a contigous area of memory which contains I*J elements.
we simulate this in C by creating an array, every element of which is an array. for example,

 Help with Code Tags 
 C++ Syntax (Toggle Plain Text) 
int a[50][30] ; // automatic or static storage class
int (*b)[30] = malloc( sizeof( int[30] ) * 50 ) ; // dynamic storage class
int a[50][30] ; // automatic or static storage class
int (*b)[30] = malloc( sizeof( int[30] ) * 50 ) ; // dynamic storage class