Strings and structures in c

Question

#include "stdafx.h"
#include <stdio.h>

struct s 
{
  char *st;
  struct s *sp; 
};

struct s *p1,*p2;
void swap(struct s *p1,struct s *p2);

int main()
{
    int i;
    struct s *p[3];
    static struct s a[]={
        {"abc",a+1},{"def",a+2},{"ghi",a}
    };
    for(i=0;i<3;i++)
    {
     p[i]=a[i].sp;
    }
    swap(*p,a);
    printf("%s %s %s\n",p[0]->st,(*p)->st,(*p)->sp->st);
    return 0;
}
void swap(struct s *p1,struct s *p2)
{
    char *temp;
    temp = p1->st;
    p1->st = p2->st;
    p2->st = temp;
}

This program outputs as abc,abc,ghi. My doubt is what does p[0]->st,(*p)->st,(*p)->sp->st outputs.we havent intialised st with abc or ghi.How does it outputs the string?

p[0] and *p are essentially the same thing.

user195488
– user195488

2011-08-10 17:20:02 +00:00
Commented Aug 10, 2011 at 17:20 — user195488
– user195488, Commented Aug 10, 2011 at 17:20
What do you mean, you haven't initialized st?

Chris Lutz
– Chris Lutz

2011-08-10 17:29:38 +00:00
Commented Aug 10, 2011 at 17:29 — Chris Lutz
– Chris Lutz, Commented Aug 10, 2011 at 17:29

Jason · Accepted Answer · 2011-08-10 17:27:17Z

4

We havent intialised st with abc or ghi. How does it outputs the string?

The value of the st member for each structure in the statically allocated array a is actually initialized through an initialization list. Writing

static struct s a[]={
    {"abc",a+1},{"def",a+2},{"ghi",a}
};

has the same effective meaning after its execution as writing the following:

static struct s a[3];
a[0].st = "abc";
a[0].sp = a+1;
a[1].st = "def";
a[1].sp = a+2;
a[2].st = "ghi";
a[2].sp = a;

And what's effectively happened after both methods of initialization is you have a statically-allocated circular linked list of struct s, where the data-members of each node in the list (the st member) is pointing to a string literal like "abc", "def", etc. The sp data member is pointing to the next node in the linked list.

answered Aug 10, 2011 at 17:27

Jason

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1 Comment

user195488 Over a year ago

This answers it. If I wasn't out of votes, I would upvote you.

sergio · Accepted Answer · 2011-08-10 17:36:32Z

2

I haven't analysed all the statements, but assignment to ->st and ->sp happens here:

static struct s a[]={
    {"abc",a+1},{"def",a+2},{"ghi",a}
};

the rest are games with pointers so the output is what you see. So, say:

the a array is created and initialized;
in the for loop the p array is also initialized;
noticing that sp recursively points to struct s instances, p and a have the same structure;
each elements of p is made point to a struct s instance taken from one of the elements of a.

edited Aug 10, 2011 at 17:36

answered Aug 10, 2011 at 17:22

sergio

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3 Comments

Angus Over a year ago

Yeah.. a+2 gets assigned to sp so moving to the 3rd location of a[] gives the st as "ghi".Thanks sergio.

sergio Over a year ago

@Code Monkey:a + n is an array position, as you say; the fact is that struct s is a recursive structure, and its sp member is made point to another struct s inside of the same array; actually, I don't see any assignment to sp in the loop... sp value is assigned to p...

user195488 Over a year ago

@sergio: I meant, p[i]=a[i].sp; - oops.

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