C programming, use of pointers

Question

I am unable to understand the following code in С, using pointers

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

int bags[5]={20,5,20,3,20};
int *next();

int main()
{
    int pos=5;
    *next()=pos;
    printf("%d,%d,%d",pos,*next(),bags[0]);
}

int *next()
{
    int i;
    for(i=0;i<5;i++)
    if(bags[i]==20)
        return(bags+i);
    printf("Error!");
}

Can anyone explain why the ans is 20,5,20.

This code invokes undefined behaviour if the if condition fails in next function. — ajay
– ajay, Commented Apr 12, 2014 at 7:25
@ajay In theory. In practice it always returns bags+i since i = 0 on he first iteration. So the loop is useless, i=0 would do. — Jens
– Jens, Commented Apr 12, 2014 at 7:27
@jens First iteration yes.. future iterations no, as the values may have changed — M.M
– M.M, Commented Apr 12, 2014 at 7:32
@MattMcNabb I don't see any future iterations in the program as is. next() is called only once and bags never modified. — Jens
– Jens, Commented Apr 12, 2014 at 9:32

LearningC · Accepted Answer · 2014-04-12 07:23:21Z

3

The output of the program will is,

5,20,5

because

if(bags[i]==20)
        return(bags+i);

returns pointer to bags[0] since bags[0]==20 and return is a pointer to it and

*next()=pos;

writes pos value to the address pointed by next() returned address, i.e. bags[0]=pos=5

answered Apr 12, 2014 at 7:23

LearningC

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1 Comment

Argha Ray Over a year ago

ya sorry the output will be 5,20,5.I was experimenting with the input data so mistakenly copied the wrong data. thanx a lot

Andrei Bozantan · Accepted Answer · 2014-04-12 07:29:11Z

0

The function next returns a pointer to the first array element equal to 20. So, *next()=pos; will modify the first element of the array from 20 to 5. Then it is simple to understand why the output is 5,20,5: since pos is five, *next() returns 20 and bags[0] is 5 as explained above.

answered Apr 12, 2014 at 7:29

Andrei Bozantan

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Comments

yasouser · Accepted Answer · 2014-04-12 07:39:38Z

The output is 5,20,5.

Note: int *next(); ==> next() is a function that returns int*.

[1] The call to next() from the line *next() = pos;. C evaluates the function from left so next() gets called first. Because of if (bags[i] == 20), the function next() returns bags+i which is simply bags since i is 0 at the moment. So next() returned the address of bags which means *next() is nothing but the pointer to bags[0]. So bags[0] becomes 5.

[2] When the execution comes to this line: printf("%d,%d,%d",pos,*next(),bags[0]);, it prints value of pos i.e 5, value of *next() which is 20 as next() returned the address of bags[0].

[3] *next() prints 20 because, this time next will return the address of bags[2] since bags[0] got assigned with the value of pos in the line before the printf().

[4] And finally, bags[0] is 5. See the explanation in [1].

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