pointer and array

general
, ,
1

#include<stdio.h>
#include<math.h>

main( )
{ static int a[ ] = { 13, 1, 2, 3, 4 } ;

int *p[ ] = { a, a + 1, a + 2, a + 3, a + 4 } ;
printf ( “\n%u %u %d”, p, *p, * ( *p ) ) ;

2

explain the output

3

I think your intention is this :-

#include < stdio.h>

#include < math.h>

int main( )
{
static int a[ ] = { 13, 1, 2, 3, 4 } ;

    int \*p[ ] = { a, a + 1, a + 2, a + 3, a + 4 } ;


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

return 0;

}

When ever you declare an array ‘temp’. If you try to get value of ‘temp’ i.e; printf("%u",temp);
It gives base address of ‘temp’ and if you tried to get ‘*temp’ then you get contents in base address of ‘a’.
So using this approach,try to understand output :

p -> base address of p,
*p -> contents in base address of p i.e base address of a,
*(*p) ->contents in base address of a i.e 13

I hope this little explanation may understand.Still if you are uncomfortable comment below

4

thanks…

5

thanks…