C Programming - Pointers - Discussion

Please explain about
p = (int*)(p+1);
printf("%d", *p);

p is a char. pointer and it is assigned to complete address of arr array. so when we write (ar+1) it take address next beyond arr array. which having value 0 of defined pointer array.

The char pointer only stores the first byte address of integer. So while increment the pointer the next value is to be printed as 0.

The answer is correct. We can convert int pointer to char pointer and viz versa. *p is apoint to first variable. *p+1 points the next array, so the answer is zero.

p is declared as a char pointer but arr is an integer array
the assignment p = arr produces a compile-time error.

p=arr;//here p is a character pointer. we should not assign int array base address...

#include<iostream>
int main()
{
using namespace std;
int a[]={1,2,3};
int *p;
p=a;
p=p+1;
cout<<*p;
return 0;
}

This will return 2, why? It is similar to that of the above question.

Can anybody explain this please ?

Assume ,

arr[0] = 1000.
arr[1] = 1004.
arr[2] = 1008.

now,

p=arr; i.e p=1000.

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

At this statement character pointer p 1st typecasted to integer pointer and again typecasted to character pointer.

printf("%d, ", *p);-prints value 2.

As memory representation for little endian (lowest byte will stored 1st).

00000010 00000000 00000000 00000000

p = (int*)(p+1);-in this statement p which is char pointer is incremented by one so it is pointing at location 1001.

printf("%d", *p);-prints value 0 because at memory location 1001 value 0 is stored (according to little endian ).

There is a warning of converting int* to char* but, there is no error. Actually we do not follow such coding practice but for the question answer is 2, 0.