Manish Bharti Manish Bharti - 1 year ago 66
C Question

not able to understand the output of this c pointer to arraycode

first of all sorry for my English .

Now come to question

here is a code about the array to pointer
problem is that i dont understand the output
according to what i have studied > *p -----means value at address p is containing
and *(*p) value at address at (value at address p congaing )
then why the output is different



void main()
static int a[ ] = { 0, 1, 2, 3, 4 } ;
int *p[ ] = { a, a + 1, a + 2, a + 3, a + 4 } ;
printf ( "\n%u %u %d", p, *p, * ( *p ) ) ;

and output is

3545106368 6293840 0

please help me to understand this output .Keep it simple and easy for me
thanks in advance...........

Answer Source

In memory your arrays looks something like this:

| a[0] | a[1] | a[2] | a[3] | a[4] |
 ^      ^      ^      ^      ^
 |      |      |      |      |
| p[0] | p[1] | p[2] | p[3] | p[4] |

I.e. each element of p is pointing to the corresponding element in a.

Then you have to learn that when using an array, it decays to a pointer to its first element, i.e. the expression p is the same as &p[0]. So when you are printing out p then you are printing out the address of the first element of the array p.

When you dereference the pointer given by p, then you are effectively doing *&p[0]. The dereference and address-of operators cancel each other out, so when you print *p you are printing p[0] which is the value of the first element in the array p.

That element, p[0], is in turn a pointer to a[0]. So when you dereference that pointer (in **p, which is the same as *p[0]) then you get the value of a[0].

Also, for any array or pointer p and index i, the expression p[i] is the same as *(p + i). From that follows that p[0] is the same as *(p + 0) which is the same as *(p) which is the same as *p.

From that follows the logic that p and &p[0] are the same.

And to really screw you around, due to the commutative property of addition, the expression *(p + i) is the same as *(i + p) which leads to the peculiar but valid expression i[p].

Last but not least, when printing a pointer with printf, you should really be using the "%p" format. It takes a void* as argument, and you also need to cast the pointers for the program to be valid. So your printf call should look like

printf ( "%p %p %d\n", (void *) p, (void *) *p, **p ) ;

See e.g. this printf (and family) reference for more information.

Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download