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Hi to everyone,
 
I have very basic doubt in Pointer.
 
I have declared a variable like this
int **value;
 
Book Definition:
value = &anotherpointer;
 
But I have used the same double pointer for storing data in two dimensional array.
 
I have given a sample code below:
              value = (int **) malloc(sizeof(int)*2);
              value[0] = (int *)malloc(sizeof(int) *2);
              value[1]= (int *)malloc(sizeof(int)*2);
    
 
This above syntax is working fine with my program.
 
Basically, i have tried to create a dynamic two dimensional array of size (2X2).
 
The above syntax of creating dynamic array works fine for all user defined and built-in data types.
 
Please let me know how memory will be allocated if i tried to create a dynamic structure pointer or other data types of size (example 10).
 
Please let me know whether i have used the pointer correctly or not?
 
And let me know the reason. Looking forward for your replies....
Posted 28-Aug-12 23:25pm
Edited 28-Aug-12 23:28pm
CPallini307.9K
v2

1 solution

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Solution 1

Yes, you may use a double pointer to store the address of another pointer as well for creating a matrix, the compiler won't complain. Smile | :)
 
The memory you are going to use does actually depend on the malloc statements.
 
Plaese note:
balasubramaniyan94 wrote:
value = (int **) malloc(sizeof(int)*2);

 
The above line is wrong, it should be
value = (int **) malloc(sizeof(int *) * 2);
 
(You are lucky if, as often happens, sizeof(int) == sizeof(int *)).
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Comments
Mihai Vrinceanu at 29-Aug-12 9:00am
   
On 32 bit operating systems, sizeof(int) is 4 and sizeof(int*) is 4.
On 64 bit operating systems, sizeof(int) is 4 and sizeof(int*) is 8.
For storing a 64 bit address, you need 8 bytes: 8 bytes * 8 bits/byte = 64 bits.

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