CODEWITHSUNDEEP
c
Hari Ram
Hari Ram

Reputation: 56

C Language and Datastructures

I have a doubt in dynamic memory allocation (malloc) Say 

ivar=(int*)malloc(1*sizeof(int));

What the above code will do? Will it create address for allocations?

Also which is the standard way to get values in malloc? (Say i as loop variable)

scanf("%d",&ivar[i]);

OR

scanf("%d",ivar+i); 

int main()
{

    int *ivar;
    ivar=(int*)malloc(1*sizeof(int));
    printf("%u",ivar); // outputs 2510
    printf("%u",&ivar);// outputs 65524

}  // please explain why it is…

Thanks in advance.

Upvotes: 0

Views: 77

Answers (1)

arvind
arvind

Reputation: 275

The memory allocated using malloc is created in heap section of RAM.

ivar=(int*)malloc(1*sizeof(int));

The syntax for malloc is 

void *malloc(size_t size);

(1*sizeof(int)) gives 4 bytes, so 4 bytes is allocated in Heap.

You cannot directly access memory of heap, so ivar pointer is used to access it.

When you write

printf("%p",ivar); // outputs 2510

printf("%p",&ivar);// outputs 65524

Both of these gives address, first one gives address of the location pointer is pointing at, second one gives address of the pointer

scanf("%d",&ivar[i]);

and

scanf("%d",ivar+i);

both are equal, so you can use either one of them.

Upvotes: 1

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