CODEWITHSUNDEEP
cmallocdynamic-memory-allocationrealloccalloc
Divya
Divya

Reputation: 393

How does realloc work on memory allocated using calloc?

When we try to resize the memory allocated by malloc using realloc, we typically do this:

char *ptr = (char *)malloc(size_1);
ptr = (char *)realloc(ptr, size_2);

If size_2 may be larger or smaller than size_1. If new size is larger then the old data is not lost and newly allocated bytes are uninitialized. The starting address contained by the ptr may change if there is not sufficient memory at the old address to store all bytes consecutively. realloc moves the contents of old block into the new block and ptr will be pointing to the initial byte of this new block.

But, if memory is allocated using calloc, I was not able to understand how realloc function acts. Can someone please give me a brief overview about how realloc works on memory allocated by calloc?

Upvotes: 1

Views: 14800

Answers (3)

ameyCU
ameyCU

Reputation: 16607

memory block layout is different for malloc and calloc

Actually no. The difference between calloc and malloc is calloc() initializes the allocated memory with 0 values, whereas malloc() doesn't initialize the allocated memory, so the memory will have undefined/garbage data.

And the number of arguments into them.

And I don't think that realloc treats memory allocated by malloc or calloc differently.

EDIT

On your problem:

a = calloc(5,sizeof(int));

if (a == NULL)
{
    printf("Error in allocating memory");
}
a = realloc(a,7); // Also check its return.

This will work as intended.

Upvotes: 3

user207421
user207421

Reputation: 310911

We know that the memory block layout is different for malloc and calloc.

No. It is the same. There is no basis for this assertion.

When we try to resize the memory allocated by malloc using realloc, we typically do this:

char *ptr=(char *)malloc(size_1); ptr=(char *)realloc(ptr, size_2);

[where] size_2 may be larger or smaller than size_1. If new size is larger then the old data is not lost and newly allocated bytes are uninitialized. The starting address contained by the ptr may change if there is not sufficient memory at the old address to store all bytes consecutively.

Correct.

realloc moves the contents of old block into the new block

If necessary. If the block was merely grown, this step isn't necessary.

and ptr will be pointing to the initial byte of this new block.

Correct.

But, if memory is allocated using calloc, i was not able to understand how realloc function acts.

The same.

Can someone please give me a brief overview abt realloc works on memory allocated by calloc?

It is the same.

You're overthinking this. calloc(n, size) is implementable with nothing more than malloc(n*size) followed by memset(). realloc() doesn't care.

Upvotes: 2

AnT stands with Russia
AnT stands with Russia

Reputation: 320491

We know that the memory block layout is different for malloc and calloc.

No, we don't. Actually, we know that there's no difference at all, aside from the fact that calloc() is responsible for multiplying its parameters (to determine the block size) and making sure that the allocated block is initialized with all-zero bit pattern.

The rest follows. There's no difference in how the memory block is treated by realloc(), regardless of what function was used to allocate it. realloc() handles calloced blocks in exactly the same way it handles malloc()ed blocks.

Upvotes: 12

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