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Question on Pointer Arithmetic
By KartalKaan, 11 years ago, In English

Hello everyone! I have two codes, they print different things but I don't know why. The first one is :

#include <stdio.h>
int main()
{
   int p[6]={1,2,3,4,5,6};
   int *q=(int *)(&p+1);
   printf("%ld\n",q-p);
   return 0;
}

And the second one :

#include <stdio.h>
int main()
{
   int p[6]={1,2,3,4,5,6};
   int *q=(int *)(p+1);
   printf("%ld\n",q-p);
   return 0;
}

First code prints 6, whereas the second one prints 1. The only difference is that the first code has an ampersand before "p+1". Do you know the reason? Thanks.

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11 years ago, # |
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The first code has an ampersand before p, not before p+1. It's high-priority unary operator, and &p+1 is equal to (&p)+1.

Basically, p+x (where p is a T* and x is an int) is equivalent to (T*)((char*)p + sizeof(*p)*x), i.e. if you add one to a pointer, it shifts by one element to high addresses.

When you take address of p, you get a pointer to an array of six integers. Size of such array is six integers, and when you add one to this pointer, you get a pointer to the byte past end of the array (six integers after the beginning).

But, if you use p as a pointer itself, it behaves like pointer to integers, and adding one to it results into pointer to the second element of an array.

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    11 years ago, # ^ |
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    Thanks for explanation. We have an important exam tomorrow which is mostly about pointers and maths. Waiting for your good wishes :)

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