C Programming - Pointers - Discussion

What is type casted?

p = (int*)(p+1) ; what it does actually ?

Why not upper byte of 2 not considered here ? (00000000).

Is it put at the end of the array ?

It gives warning as incompatible pointer type.

Answer 2, 0 is correct.

It doesn't show errors.

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.

In borland and turbo c++ it gives the error: Cannot Convert "int*" to "char*".

since integer requires 4 bytes=32 bits so 2,3,4 stored as
==>{lsb first then msb}
00000000 00000000 00000000 00000010 00000000 00000000 00000000
103 102 101 100 107 106 105
00000011 00000000 00000000 00000000 00000100
104 111 110 109 108

Initially,p holds address=100
So value at addr 100=2.

Now,it casted to char ptr,
So int*(p+1) =value at addr 100+1=0.

It causes a compile time error as it is an incompatible type conversion p being an character pointer is referring to int base array.

It causes a compile time error as it is an incompatible type conversion p being an character pointer is referring to int base array.

Anyone please explain the correct logic.