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.

You mislabeled your columns: it should be *p b a, not a b *p. — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Mar 5, 2013 at 13:26
For the sake of change-resistence, better do int b[sizeof(a)/sizeof(a[0])] = {... and replace the 5 values by the same term. This way, you can change the array a and everything will be ok. — glglgl
– glglgl, Commented Jul 12, 2013 at 6:24

Lara Bailey · Accepted Answer · 2013-03-05 13:24:50Z

9

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.

answered Mar 5, 2013 at 13:24

Lara Bailey

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3 Comments

user2135885 Over a year ago

so basically it can be anything, just in this case it appears to point to the next address in the memory in this case which it points to the starting element of array a? correct me if i am wrong. Thanks

UmNyobe Over a year ago

exactly. it can be anything. The best case is that it crashes so you are aware something is wrong early.

Lara Bailey Over a year ago

@user2135885: To me it seems like a reasonable explanation although it is possible that there is a more obscure reason for the coincidence that you are seeing.

hankym · Accepted Answer · 2013-03-05 14:46:37Z

1

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].

answered Mar 5, 2013 at 14:46

hankym

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Comments

Yu Hao · Accepted Answer · 2013-07-12 06:10:38Z

1

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;
}

edited Jul 12, 2013 at 6:10

Yu Hao

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answered Jul 12, 2013 at 6:07

user1998844

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1 Comment

glglgl Over a year ago

better: p = b again, as the 5 might change.

Rahul Subedi · Accepted Answer · 2013-03-05 14:50:15Z

-2

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;
}

edited Mar 5, 2013 at 14:50

answered Mar 5, 2013 at 14:44

Rahul Subedi

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