Question

What is the output of the following C code

 

int main( )
{
    int p[5] = {10, 11, 12, 13, 14};
    int *ptr = (int *) (&p + 1);
    printf(“%d %d”, *(p + 1), *(ptr – 1));
}

Answer 1

Type of pointer matters while doing pointer airthmatic.
Here, $p$ is an array of $5$ integers, although &p, p or &p[0] give same result i.e., base addresses of array (or) address of $1^{st}$ element when printed, but they behave differently when airthmatic is performed.

int p[5] = {10, 11, 12, 13, 14}; // p is an array of 5 ints 

Let array be stored like in the picture below :

printf("%d %d %d",p,&p,&p[0]); // all print same address (200)

But when pointer airthematic is performed on these, then $\Rightarrow$

p + 1 = 204 // increment by an element 
&p[0] + 1 = 204 // increment by an element 
&p + 1 = 220 // increment by sizeof(array)

//Code has been modified a bit to make point clear
int main() {
int p[5] = {10, 11, 12, 13, 14}; // let p = 200
int *ptr = (int *)(&p + 1); // ptr stores 220
int *ptr1 = (int *)(p + 1); // ptr1 stores 204
printf("%d %d %d",*(p + 1),*(ptr- 1),*(ptr1-1));
// Output : 11 14 10
return 0;
}

$\color{navy}{*(p+1) = *(204) = 11}$
$ \color{navy}{ *(ptr - 1) = *(220 -1*sizeof(int)) = *(216) = 14}$ // ptr is a pointer to int
$ \color{navy}{ *(ptr1 - 1) = *(204 - 1*sizeof(int)) = *(200) = 10}$ // ptr1 is a pointer to int

Comments

That's a really nice explanation .

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my pleasure bro.

Answer 2

p is an array of 5 int.

&p is having type pointer to an array of type int.

&p + 1 = (int)&p + 1*sizeof (p)

= (int)&p + 1* 5*sizeof(int)

= address of 6th int from start of p.

p + 1 = (int)p+ 1*sizeof(*p)

= (int)p + 1*sizeof(int)

= address of 2nd int from start of p.

So, (*p+1) = 11, *(ptr-1) = 14.

PS: Assigning an address one after an array bound is valid in C- but not dereferencing it. Also p and &p have same value (but not type) for an array p.