cant figure the output..

[nichas]

the code is ...
int main(void){
static int a[]={1,2,3,4,5};
static int *p[5]={a,a+1,a+2,a+3 ,a+4};
int i;
for(i=0;i<5;i++ ){ printf("%u\n",a +i);}
printf("\n%u\t% u\t%d",p,*p,*(* p));
return 0;
}
Now in this code the output which iam getting is :::
the address of p is 4 added to the address of element 5 in the array
'a'. That means p points to the address of the memory location next to
a[4] (since on my machine int is 4 bytes). But what i guessed is that
it would point same as *p that is the element a[0]. I am not able to
think how this can happen.. please reply where i am wrong in my
thinking.

[S.Tobias]

nichas <sa***********@ gmail.com> wrote:

the code is ...
#include <stdio.h> /* printf */ int main(void){
static int a[]={1,2,3,4,5};
static int *p[5]={a,a+1,a+2,a+3 ,a+4};
int i;
for(i=0;i<5;i++ ){ printf("%u\n",a +i);} printf("%x\n", (void*)(a+i)) printf("\n%u\t% u\t%d",p,*p,*(* p)); printf("\n%x\t% x\t%d\n", (void*)p, (void*)*p, *(*p)) return 0;
}
Now in this code the output which iam getting is :::
You should have included your output. Mine was:

-|8049490
-|8049494
-|8049498
-|804949c
-|80494a0
-|
-|80494a4 8049490 1
the address of p is 4 added to the address of element 5 in the array
'a'. That means p points to the address of the memory location next to
a[4] (since on my machine int is 4 bytes).
In my output it is the same. But this is, of course, a coincidence,
the address might be different.

`p' is an array (of ptrs to int), it is not a pointer, therefore
it doesn't point to anything. In the expression it decays to "a
pointer to the first element", therefore is also the address of
the second array, which happens to be adjacent to the first one
(and this is the reason for the difference of 4).
But what i guessed is that
it would point same as *p that is the element a[0].

In `*p', again `p' decays into "the pointer to the first element";
`*' and `p' together then yield an lvalue for the first element
(or simply `p[0]'). The first element is `a'; the identifier
denotes the array, but in the expression it decays into a pointer
to its first element (the value of `a' is also the address of the
first array). So `*p' is a pointer to the first element of `a' (&a[0])
(the first number in the output and the second one in the last line
are the same).

It's easy now to guess that `**p' is the same as `a[0]' (the last
output number is "1").

To sum up:

p == &p[0]
*p == p[0] == a == &a[0]
**p == a[0]

--
Stan Tobias
mailx `echo si***@FamOuS.Be dBuG.pAlS.INVALID | sed s/[[:upper:]]//g`

[Martin Ambuhl]

nichas wrote:

the code is ...
int main(void){
static int a[]={1,2,3,4,5};
static int *p[5]={a,a+1,a+2,a+3 ,a+4};
int i;
for(i=0;i<5;i++ ){ printf("%u\n",a +i);}
printf("\n%u\t% u\t%d",p,*p,*(* p));
return 0;
}
Now in this code the output which iam getting is :::
the address of p is 4 added to the address of element 5 in the array
'a'. That means p points to the address of the memory location next to
a[4] (since on my machine int is 4 bytes). But what i guessed is that
it would point same as *p that is the element a[0]. I am not able to
think how this can happen.. please reply where i am wrong in my
thinking.

You need to check your textbook for
1) required headers for variadic functions (e.g. printf)
2) pointer arithmetic
3) proper specifiers for pointers with printf.

Check the following code, that differs from yours in important ways, and
see if it, along with your textbook, helps you understand what is happening:

#include <stdio.h>
int main(void)
{
static int a[] = { 1, 2, 3, 4, 5 };
static int *p[5] = { a, a + 1, a + 2, a + 3, a + 4 };
int i;
for (i = 0; i < 5; i++)
printf("a + %d = %p, &a[%d] = %p, p[%d] = %p\n", i,
(void *) (a + i), i, (void *) &a[i], i, (void *) p[i]);
printf("p = %p, *p = %p, p[0] = %p, **p = %d, *p[0] = %d\n",
(void *) p, (void *) *p, (void *) p[0], **p, *p[0]);
return 0;
}
a + 0 = b6f4, &a[0] = b6f4, p[0] = b6f4
a + 1 = b6f8, &a[1] = b6f8, p[1] = b6f8
a + 2 = b6fc, &a[2] = b6fc, p[2] = b6fc
a + 3 = b700, &a[3] = b700, p[3] = b700
a + 4 = b704, &a[4] = b704, p[4] = b704
p = b6e0, *p = b6f4, p[0] = b6f4, **p = 1, *p[0] = 1

[nichas]

>

#include <stdio.h>
int main(void)
{
static int a[] = { 1, 2, 3, 4, 5 };
static int *p[5] = { a, a + 1, a + 2, a + 3, a + 4 };
int i;
for (i = 0; i < 5; i++)
printf("a + %d = %p, &a[%d] = %p, p[%d] = %p\n", i,
(void *) (a + i), i, (void *) &a[i], i, (void *) p[i]); printf("p = %p, *p = %p, p[0] = %p, **p = %d, *p[0] = %d\n",
(void *) p, (void *) *p, (void *) p[0], **p, *p[0]); Can u tell me why we need to cast p, *p to the void pointers...
One more thing where is p pointing to. Its address in ur program
is b6e0 which is before a[0]. Is there any logic or p can point
anywhere.. please reply.
return 0;
}
a + 0 = b6f4, &a[0] = b6f4, p[0] = b6f4
a + 1 = b6f8, &a[1] = b6f8, p[1] = b6f8
a + 2 = b6fc, &a[2] = b6fc, p[2] = b6fc
a + 3 = b700, &a[3] = b700, p[3] = b700
a + 4 = b704, &a[4] = b704, p[4] = b704
p = b6e0, *p = b6f4, p[0] = b6f4, **p = 1, *p[0] = 1

[Jens.Toerring]

nichas <sa***********@ gmail.com> wrote:

#include <stdio.h>
int main(void)
{
static int a[] = { 1, 2, 3, 4, 5 };
static int *p[5] = { a, a + 1, a + 2, a + 3, a + 4 };
int i;
for (i = 0; i < 5; i++)
printf("a + %d = %p, &a[%d] = %p, p[%d] = %p\n", i,
(void *) (a + i), i, (void *) &a[i], i, (void *) p[i]);

printf("p = %p, *p = %p, p[0] = %p, **p = %d, *p[0] = %d\n",
(void *) p, (void *) *p, (void *) p[0], **p, *p[0]);

Can u tell me why we need to cast p, *p to the void pointers...

Because the "%p" format specifier expects a void pointer according to
the C standard.
One more thing where is p pointing to. Its address in ur program
is b6e0 which is before a[0]. Is there any logic or p can point
anywhere.. please reply.

If you use 'p' in value context it's taken to mean the address of the
first element of the array 'p', i.e. the address of p[0], and not the
content of what's stored in p[0], which would be the address of a[0].

Regards, Jens

PS: Could you please try to avoid these "u" and "ur" abbreviations?
They make things harder to read for non-native speakers like me
and it shouldn't take you too much time to type "you" or "you're"
instead. Thank you.
--
\ Jens Thoms Toerring ___ Je***********@p hysik.fu-berlin.de
\______________ ____________ http://www.toerring.de