An interesting discussion about malloc( ) and calloc( )

What are malloc( ) and calloc( )?

Simply putting they are the predefined functions in C language.

Malloc( ) and calloc( ) are two such functions which are used for more or less identical purpose and that is to allocate memories for any variable dynamically during execution time of the program.

Give me a brief of malloc( ).

So as I have said malloc is a function; malloc must have three valid components as all other functions do have. These are (a) function name (b) function arguments and (c) function return type, that is, malloc must have a signature or prototype. Now what is the prototype of malloc? It is like following:

(void*) malloc(unsigned int) [Note: The actual prototype is slightly different than this, but for the time being this simplification will work]

It shows that the function name is malloc itself, it expects an unsigned integer as argument and it returns a void pointer or generic pointer.  So let us discuss a bit about them.

As per the prototype, malloc expects an unsigned integer. To assimilate this, let us recall why should we use malloc? Malloc is used to allocate a chunk of memory at program run time dynamically. So we must let malloc know what the size of that chunk of memory is. Suppose we want malloc to allocate a chunk of memory where we can store 10 integers. In this case, we actually want malloc to allocate 20 bytes of memory (if we are running our program on windows). And we do it by supplying the size 20 as an argument. So the unsigned integer that is passed to malloc is actually the requirement of memory in bytes; and it is evident that the memory requirement cannot be negative any time, so the data type is unsigned.     

Now what that void pointer is doing at the beginning as return type? To understand this, we have to recall that a function gives a feedback to its caller through the return type. So after allocating the memory (I admit, you do not how) malloc has to return the base pointer, I,e, the address of the starting location of the memory chunk allocated. Though void pointer, malloc actually returns this base address.

I am sure you will scream out “why void?” Wait a minute. Answer me, who has written this malloc function? Is it you or someone else? Obviously it is not you, malloc function is written by someone else. Now the original developer of malloc had no idea for which purpose you would be using this malloc. Currently, you can use malloc to allocate memory for storing some integers, or some floats or some structures defined by you. There are so many options and those options are equally likely for you. You as a programmer currently know that this malloc is being used for storing some integers or whatever else, but the original programmer had no idea about that. So malloc returns void pointer or generic pointer, a special one which can point to any type of memory generically. So it is your responsibility to properly type cast this void pointer. If you are using it to store some integers, cast void* as int*, if you are using malloc to store some Node type structures, then cast it to Struct Node*. Got it?

One last point still exists. You must have seen malloc argument to contain sizeof(), right? Why it is so?

As I have mentioned that the memory requirement is informed through the argument, it is highly possible that the memory requirement may change as the OS changes. For example in one system you may require 20 bytes to store 10 integers, while it may be 40 bytes in other system or OS. So if I hardcode my memory requirement as 20 or 40 for any specific platform, it won’t suffice the scenarios generated by other systems. So to meet the portability issue, that may rise up later, we as a programmer should not deal with the memory requirement directly. We should inform the compiler that  that we need to store n number of elements only and let the compiler decide what the size of each such element is by using sizeof.

So we should call malloc with an argument (n * sizeof(int)) if we want to store n number of integers. By doing this, we are actually letting the C compiler to calculate the memory requirement (under the current OS) of one element and multiplying this with n, compiler can compute the exact memory requirement as per the current system.

As a good programmer you should never hardcode your memory requirement, it may end up in a serious problem if your program gets ported to any other OS.

Look at the following example for clear undestanding:

To store 15 characters, do like

char *pointerToCharArray;

int size = 15;

pointerToCharArray = (char*)malloc(size*sizeof(char));

To store n numbers of MyStructure type of structure, do like

struct MyStructure *pointerToMyStructureArray;

pointerToMyStructureArray = (struct MyStructure*)malloc(n*sizeof(struct MyStructure));

To store k numbers of integers, do like

int* point;

point = (int*)malloc(k*sizeof(int));

Once you are done with these, you can use your pointer (like pointerToCharArray or pointerToMyStructureArray or point as seen in the examples) like an array with indexing. That means now point[5] will give you the integer stored at 5^th index of point or the sixth integer stored within the memory chunk, base address of which, is being pointed by a pointer point.

Huh! I finish it off here!    

So what is calloc then? How it is different from malloc?

Calloc is another function which, like malloc, is used for dynamic memory allocations.

Malloc if called to allocate a memory space that can store 10 integers (under windows),occupies the memory as a single chunk of 20 bytes, where calloc, if called for the same purpose, occupies 20 bytes, but not as a single chunk of memory, rather , it occupies 10 blocks of memory, each having 2 bytes (as it is under windows, as we have supposed). Hence calloc has the following prototype:

(void*) calloc(unsigned int, unsigned int)

The first unsigned int is for number of elements that you want to store and the second unsigned int is for memory allocation needed for storing a single element.

This is why calloc can occupy the memory space with multiple blocks, each of equal size. The void* is doing the same thing as that of malloc

So look at the following example for clear understanding:

To store 15 characters, do like

char *pointerToCharArray;

int size = 15;

pointerToCharArray = (char*)calloc(size, sizeof(char));

To store n numbers of MyStructure type of structure, do like

struct MyStructure *pointerToMyStructureArray;

pointerToMyStructureArray = (struct MyStructure*)calloc(n, sizeof(struct MyStructure));

To store k numbers of integers, do like

int* point;

point = (int*)calloc(k, sizeof(int));

So the syntactical difference is malloc takes one argument, while calloc takes two. But you need to have an idea why it is so. I think the above discussion tells you that.

Another point of difference is like following:

Malloc allocates the memory chunk and initializes the memory with garbage values, whereas calloc allocates the multiple block of memory chunks and initializes those with 0s (zeros)

So we come to an end of this discussion. I am open to clear your doubts if any.