C programming

Question

#include<stdio.h>

int main()
{
    int arr[3] = {2, 3, 4};
    char * p;
    p = arr;
    p =(char *)(int*)(p);
   printf("%d, ", *p);
    p = (int*)(p+2);
    printf("%d", *p);

    return 0;
}

Pls explain the output and also the code

Comments

it should be 2,0.

the entire array will have 12 bytes space as each integer will occupy 4B.

and pointer will take 1B for every case.

p is a pointer that will initially store arr[0] 0th cell address. which contains 2 as (00000010) .

now its address is typecasted but space / memory wont change for integer to character so it will be 1B only.

so it prints 2.

now p+2 will goto 3rd cell of arr[3] as its an pointer which has 12 cells as 12 bits allocated.

so it initially conatins 0 so 0 printed.

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i think the output depends upon the size of int we take.

Answer 1

NOTE: Output will vary for different size of int.

Assuming that integer requires 2 bytes 

p = (char*)((int*)(p));

 till now the pointer p is type casted to store the variable of type character right?

This is wrong, Type Casting donot change the Size of pointer. 

means if it would have been p=((int*)p); instead of  p = (char*)((int*)(p)); result would be same.

printf("%d, ", *p); // output : 2

integer array will be stored in memory MSB to LSB as

00000000	00000100	00000000	00000011	00000000	00000010
4		3		2

so p+2 points to that location  00000011.

printf("%d", *p); // output : 3

so answer is 2 3

Comments

You are right, but why do we use type casting on pointer if it don't make any change to its size

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it is not implementation... it is a question... he may gives for confusing us

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Ok, sir I will update answer with this statement

Answer 2

its answer 2,3 because if type casting occur then in this case 

in case of  fisrt *p then it catch only on byte due to char type  and in general computer big indian format is aaply so first is LSB=2 then msb=0 hence lsb value 2

and other case means int*(p+2 )then in this case access 1 byte due to  it access 3 aacording to big indian format