program output of a c program

Question

please explain the output of the below mentioned program. I thought that output should be 1 but it is not 1

#include <stdio.h>

int main()
{
    int  a[] = {1, 2, 3, 4, 5, 6};
    int *ptr = (int*)(&a+1);

    printf("%d ", *(ptr-1));
    getchar(); 

    return 0;
}

You may want to include more information, including the actual output you received. — Matt
– Matt, Commented Apr 15, 2015 at 19:32
type of &a is int (*)[6]. it meant pointer to int[6]. so &a + 1 point to next(a[6]) the last element(6:a[5]). then cast to int * then -1 so point to last element 6. — BLUEPIXY
– BLUEPIXY, Commented Apr 15, 2015 at 19:41

Iharob Al Asimi · Accepted Answer · 2015-04-15 19:47:46Z

3

It's because &a has type int(*)[6] and hence &a + 1 != &a[0] + 1 which is what you meant, given your expected output.

If you print the addresses &a and &a[0], they will be the same, however incrementing &a is not the same as incrementing &a[0] because sizeof(int) != sizeof(int *), although that's not always true, your code should not assume it is, or depend on that.

Also, it appears that your are using the cast because your compiler was complaining about the fact that &a is of an incompatible pointer type, you should not need the cast if the pointers are of compatible types so

#include <stdio.h>

int main()
{
    int  a[] = {1, 2, 3, 4, 5, 6};
    int *ptr = &a[0] + 1;

    printf("%d\n", *(ptr-1));

    return 0;
}

should compile just find, and the output will be 1 as you expected.

edited Apr 15, 2015 at 19:47

answered Apr 15, 2015 at 19:32

Iharob Al Asimi

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4 Comments

Iharob Al Asimi Over a year ago

The downvoter seems to disagree, I would like to see your answer to see if mine is really wrong or bad.

dev_k2.0 Over a year ago

i am having the confusion around the same thing.. can you please elucidate it little more

Alex Reynolds Over a year ago

Not sure why this was voted down. Would be interested to see the downvoter comment on their reasoning.

Keith Thompson Over a year ago

More precisely, &a has type int(*)[6] (now fixed, thanks). Also, sizeof(int) != sizeof(int *) -- sizeof (int*) is irrelevant. What's relevant is sizeof (int[6]).

Loic · Accepted Answer · 2015-04-15 20:35:50Z

1

a and &a are the same value,
a is of type int [6],
&a is of type int (*)[6].

It means that

    ptr
 == (int *)(          &a + 1                )
 == (int *)( (char *) &a + 1*sizeof(*&a)    )
 == (int *)( (char *) &a + 1*sizeof(a)      )
 == (int *)( (char *) &a + 1*6*sizeof(int)  )
 == (int *)( (char *) a  + 1*6*sizeof(int)  )
 == (int *)( (char *) a  + 1*6*sizeof(*a)   )
 == (int *)(          a  + 6                )
 == &a[6]

thus

*(ptr-1) == *(a+5) == a[5] == 6

the reason why "6 " is printed.

answered Apr 15, 2015 at 20:35

Loic

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