C Programming - Pointers - Discussion

Here the memory is allocated by malloc function. P is local variable so the memory is allocated on stack for P. But the memory allocated by malloc function is 100 byte and which is allocated on HEAP. Here the type of pointer is character so for allocating the memory malloc is type casted with character, Because malloc always returns void*.

So void* is a pointer which pointing to particular address but can not access the element from that address. For allocating memory malloc internally call sbrk and brk algorithm. This algorithm internally called by growrage algorithm.

#include<stdio.h>
int main(){
int i=320;
char *ptr=(char*)&i;
printf("%d",*ptr);
return 0;
}

Ans: 64.

Could anyone explain this program and answer?.

1. char p;
p= (char *) malloc (1000) ;

2. char *p;
p= (char *) malloc (1000) ;

Is that 1 and 2 mean the same. explain?

malloc & calloc both are memory allocation function & both returns void type where afterwords we type case it.

ex:

int *p;
p=(int*)malloc(1000);

where malloc depends upon byte which are fixed i.e 8-bit,again 16-bit.

calloc deals with no.of bytes where bytes is considered as block relevant to the memory issue.

I want to explain brief discription about the char that char represents a individual value during coding in c language i.e. p in above code,as we can also write as char p = malloc(100);

Explanations are good.

But one think is missing that since *p points to the beginning address, thus p=char* malloc(100), can be assigned with a char*p.

The type should be same is the necessary point but why so works is because of p(base address).

@Vij.

Both are not the same.

Because
1) not a proper variable pointer declaration.
So, 2) proper define the pointer variable.

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).

Not getting it, Explain it in detail.

Explain it in detail.