C Programming - Memory Allocation - Discussion

@Cherry.

It's right.

@All.

ptr is a char ptr which does not hold any address it's a const ptr so when we allocate memory with ie, ptr =(char *) malloc(30); after it stores the RAM in it. bt when we use lik dis i,e *ptr = (char ) malloc (30); its a segmentation fault(core dumped).

Because first of all its typecasting is wrong and second one is ptr is already pointer bt again they use ptr as *ptr. so for correct answer we must use ptr = (char *)malloc (30);.

If I made any mistake correct me. Thank you.

What about strcpy(ptr, "RAM");?

ptr stores the address? Is this statement right or not?

malloc return value is void *, when write only (char)malloc(); then get error or segmentation fault but, write (char*) malloc();

Then write proper value RAM.

malloc return void type and so we have to type cast into type of the LHS. ptr is the pointer to the char type. So we should use char*.

It's a wild pointer. Because, ptr is declared but not defined (that is not pointed to any char variable). So, it's pointing to a garbage address and we are trying to write the value in an unallocated memory. It leads to wild pointer.

It should be

char *ptr = (char *)malloc(30);

or

char *ptr;
ptr = (char*)malloc(30);

Then the output of the program will be "RAM".

Can't we say that malloc(30) wil return address of allocated memory (char)malloc(30) will convert it to char and then *ptr can hold it.

int example=1;
char *p=(char)example ; (is it equivalent to char *p="1"; )

I am confused. Please help ?

Address of allocated memory by using malloc should be stored in ptr i.e. pointer to an character. Then only it will work perfectly.

It is not a character conversion it should be pointer to character when returned from malloc().