C Programming - Pointers - Discussion

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

This should give the output as 3 because %d access 2 bytes of memory so even if we start accessing the memory from 2nd byte from the left, answer should be 3 i.e. 00000000 00000011.

It all depends on compiler as well.

First p = (int *)(p+1); is a error, reason is that p is char pointer so if some of you guys are getting value of *p its because it didn't actually converted to int pointer, your compiler failed to convert and instead of raising error it just raised warning.

I tried in Turbo C and my observation is if array starts from 1000.

1000 -> 2.
1001 -> 0.
1002 -> 3.
1003 -> 0.
1004 -> 4.

Now, this statement p = (int*)(p+1) in the problem should have been written as:

1. p = p+1;

//Now p starts from 1001.

OR.

2. int*q = (int *)(p+1);

//Now q starts from 1001.

Consider Case 1: printf ("%d", *p); //Output: 0.

Consider Case 2: printf ("%d", *q); //Output: 768.

Reason: q is an int pointer so it has to access 2 bytes.

Remember this is little endian architecture so, 1002 byte address will get accessed first and then 1001.

Therefore the value would be represented in q as following.

00000011 00000000.
1002 1001.
This amounts to 768.

Note: If in your system int is 4 bytes then value would be different but it won't be 0.

And those who are saying that p = (int *)(p+1); works, p is not actually getting converted to int pointer, so the answer is of no significance.

Nice explanation @Puru.

It will result in error, can't convert (int*) to (char*).

What is meant by an endian method?

Those who are getting an error.

In some compiler, you can equate two different data type pointers so you have to typecast them.

Let's say the base address of the array is 200.

p = (char*)(arr+1);

Here p will be 204.

If, p = (char*) arr+1;

Then it will be 201.

Very good and clear explanation. Thank you all.

Do we need to typecast the pointer every time?

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

Why do this type casting?

Type casting means?