how to allocate memory using malloc in a 3 D array

[jmbn2k]

Hi
Can anyone pls teme how to allocate memory using malloc in a 3 D diemensional array.............

[tyreld]

Take a look at this thread:

http://www.thescripts.com/forum/thread546378.html

[jmbn2k]

Take a look at this thread:

http://www.thescripts.com/forum/thread546378.html

HI
Can yo plz get me d full code coz dis was an intw question which am supposd to answer as soon as possible......as am a fresher i don hav much knowledge in dis conept plz......help me wit the code n the explanation of d code plz.......thank you so much for your previous reply

[vermarajeev]

int main(void)
{ 
    int*** arr;
    arr = new int**[2];
    for(int i=0; i<5; i++)
    {
        arr[i] = new int*[5];
        for(int j=0; j<3; j++)
        {
            arr[i][j] = new int[2];
        }
    }
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                arr[i][j][k]=2;
            }            
        }    
    }
 
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                cout<<arr[i][j][k]<<"\t";
            }
            cout<<"\n";
        }
        cout<<"\n";
    }
 return 0;
}

[jmbn2k]

int main(void)
{ 
    int*** arr;
    arr = new int**[2];
    for(int i=0; i<5; i++)
    {
        arr[i] = new int*[5];
        for(int j=0; j<3; j++)
        {
            arr[i][j] = new int[2];
        }
    }
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                arr[i][j][k]=2;
            }            
        }    
    }
 
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                cout<<arr[i][j][k]<<"\t";
            }
            cout<<"\n";
        }
        cout<<"\n";
    }
 return 0;
}

thank you so much.......

[tyreld]

int main(void)
{ 
    int*** arr;
    arr = new int**[2];
    for(int i=0; i<5; i++)
    {
        arr[i] = new int*[5];
        for(int j=0; j<3; j++)
        {
            arr[i][j] = new int[2];
        }
    }
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                arr[i][j][k]=2;
            }            
        }    
    }
 
 
    for(int i=0; i<5; i++)
    {
        for(int j=0; j<3; j++)
        {
            for(int k=0; k<2; k++)
            {
                cout<<arr[i][j][k]<<"\t";
            }
            cout<<"\n";
        }
        cout<<"\n";
    }
 return 0;
}

This code has failed to answer the question. First, the question asked about using "malloc" not the "new" operator. Second, you have provided faulty code demonstrating that you yourself do not understand the concept. The code you have provided exceeds the bounds of the array you define. The first example being that your first dimension is intialized to hold 2 elements yet you use a bounds of 5 when trying to allocate the second dimension.

[tyreld]

The original thread I referenced should have contained enough information to answer your question. Here is the code again including an extra for loop that initializes each value of the array to the product of its indices.

#include <stdlib.h>
 
#define X_SIZE 5
#define Y_SIZE 4
#define Z_SIZE 3
 
int main(void)
{
   int ***array;
   int i, j, k;
 
   array = (int ***)malloc(sizeof(int **) * X_SIZE);
 
   for (i = 0 ;  i < X_SIZE; i++) {
      array[i] = (int **)malloc(sizeof(int *) * Y_SIZE);
 
      for (j = 0; j < Y_SIZE; j++) {
         array[i][j] = (int *)malloc(sizeof(int) * Z_SIZE);
 
          for (k = 0; k < Z_SIZE; k++)
             array[i][j][k] = i * j * k;
      }
   }
 
   // Do Some Stuff With Your 3D Array
 
   return 0;
}
 

[jmbn2k]

The original thread I referenced should have contained enough information to answer your question. Here is the code again including an extra for loop that initializes each value of the array to the product of its indices.

#include <stdlib.h>
 
#define X_SIZE 5
#define Y_SIZE 4
#define Z_SIZE 3
 
int main(void)
{
   int ***array;
   int i, j, k;
 
   array = (int ***)malloc(sizeof(int **) * X_SIZE);
 
   for (i = 0 ;  i < X_SIZE; i++) {
      array[i] = (int **)malloc(sizeof(int *) * Y_SIZE);
 
      for (j = 0; j < Y_SIZE; j++) {
         array[i][j] = (int *)malloc(sizeof(int) * Z_SIZE);
 
          for (k = 0; k < Z_SIZE; k++)
             array[i][j][k] = i * j * k;
      }
   }
 
   // Do Some Stuff With Your 3D Array
 
   return 0;
}
 

thanx a lot........actually i dont hav much idea on dis dats y i asked for clear explanation n now its clear........

[vermarajeev]

This code has failed to answer the question. First, the question asked about using "malloc" not the "new" operator. Second, you have provided faulty code demonstrating that you yourself do not understand the concept. The code you have provided exceeds the bounds of the array you define. The first example being that your first dimension is intialized to hold 2 elements yet you use a bounds of 5 when trying to allocate the second dimension.

Hi tyreId,
Thankx for your comments,
I'm accept my mistake....
I think I was unsure about what I wrote

Also I just provided solution using new coz the thread link already contains solution using malloc and I thought to implement using new....

Never mind thankx for your comments and I think here is the solution you were looking for

int main(void)
{
   int ***p;
   p = new int**[5];
   for(int i=0; i<5; ++i)
   {
    p[i] = new int*[5];
    for(int j=0; j<5; ++j)
    {
         p[i][j] = new int[5];
    }
   }
   return 0;
}