C Programming - Pointers - Discussion

Why are we using return 0 if we give the main function as int main? can anyone tell me?

In C, 0 = true and others values represent false; then when one return 0 is to say that no errors occur.

Explain it in detail.

Not getting it, Explain it in detail.

step 1: char p // It is normal.
step 2: here we are using pointer so p = p* and p = 1000(size).
step 3 : P* = (char*)malloc(1000).

Malloc is a function to allocate memory dynamically(i.e. at the runtime of a program). So, when the memory being declared by the compiler at runtime it doesn't know the type of memory being allocated because we can't specify the data type of memory at runtime like static variables. hence we have to typecast that memory according to our need. (In this case as Character type).

Hope it helps.

The pointer declaration and initialization can be done in two ways.

int *ptr ;
ptr = &a;
or
int *ptr =&a;

so
char *p;
p = (char*) malloc(100);
It can also be written as;
char *p = (char*)malloc(100);

The malloc function creates a memory in heap memory and returns void pointer. which can be converted into any pointer type. now so char *p=(char *)malloc(sizeof(int)*10);

Nowhere,

size(int)*10 means want to create 20 bytes of memory.
malloc(sizeof(int)*10) means 20 bytes of memory created by malloc function in heap memory.
now malloc function returns void pointer. A void pointer can convert into any other pointer type.
(char *)malloc(sizeof(int)*n)

Here we had converted the void pointer into char pointer.

That's it.

Thank you all for your good explanations.

Can anyone explain what is the relation between arrays & pointers?

Prototype of malloc = ptr = (data type *)malloc(size);