c pointers and array

Question

#include <stdlib.h>
#include <stdio.h>

int main (void)
{
    int a[] = {1,2,3,4,5};
    int b[] = {0,0,0,0,0};
    int *p = b;

    for (int i =0; i < 5; i++)
    {
        b[i] = a[i]+1;
        *p = a[i]+1;
        p++;
    }
    for (int i = 0; i < 5; i++)
    {
        printf (" %i \t %i \t %i \n", *p++, b[i], a[i]);
    }
    return 0;
}

For this code I get why the output for a and b but why does the pointer have the same value of a?

*p is b[0] = a[0]+1, isn't it? So p++ means next address over for b so it's b[1]=a[1]+1.

ie 
*p  b  a
1  2  1
2  3  2
3  4  3
4  5  4
5  6  5

Answer 1

You are getting undefined behavior. At the end of the first loop p points to "one past the end" of b. Without resetting it, you then dereference it and continue to increment it, both of which cause undefined behavior.

It may be that on your implementation the array a is stored immediately after array b and that p has started to point into array a. This would be one possible "undefined" bahaviour.

Answer 2

after the first for{},p point at b[5],but the size of b is 5,so b[5] value is unknow,the printf *p is the same value as a[i],the reason may be in memory b[5] is a[0].

Answer 3

I think what you need to do is to add a p = p - 5;

#include <stdio.h>
int main (void)
{
    int a[] = {1,2,3,4,5};
    int b[] = {0,0,0,0,0};
    int *p = b;
       int i =0;

    for (i =0; i < 5; i++)
    {
        b[i] = a[i]+1;
        *p = a[i]+1;
        p++;
    }
    p = p - 5;
    for (i = 0; i < 5; i++)
    {
        printf (" %i \t %i \t %i \n", *p++, b[i], a[i]);
    }
    return 0;
}