c pointers and array

Question

#include 
#include 

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 <         


        

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

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;
}

Answer 3

you shouldnt make seperate loop for printing and incrementing the values of the array . do both in same loop and do the follouwing to get ur output :) #include #include

int main(void)
{

int a[]={1,2,3,4,5};
int b[]={0,0,0,0,0};
int c[]={0,0,0,0,0};
int *p;
int i;
p=c;
for( i =0 ; i<5 ; i++)
{
    b[i]=a[i]+1;
    *p=b[i]-1;
    //*p++;

//for( i =0 ; i<5 ; i++)

    printf(" %i \t %i \t %i \n" ,*p,b[i],a[i]);

}
return 0;
}

Answer 4

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].