Answer Posted / anag

*(arr+3)------>arr[0][3] that means the there is any chnage
in the last value of an array
{1,2,3,--}

we know ++ has higher prededence than * so
*++ptr---->*(++ptr)
*(++ptr)----> increment in the location after that it point
to the value
it represent the second location of an array
* represent the value at this address
the value at the second location is 2.
in the second expression first it refer the value after
that it increment in the location
ptr currently points to the second location . ptr holds
that location for the second expression * represent the
value at that location that is 2.
so 2+2->4
{1,2,3,4} ----------->ans
suppose if we add a another expression after this that *ptr
then it print the value 3
because previous expression increment the location of the
value
Thank you

main() { struct s1 { char *str; struct s1 *ptr; }; static struct s1 arr[] = { {"Hyderabad",arr+1}, {"Bangalore",arr+2}, {"Delhi",arr} }; struct s1 *p[3]; int i; < BR> for(i=0;i<=2;i++) p[i] = arr[i].ptr; printf("%s ",(*p)->str); printf("%s ",(++*p)->str); printf("%s ",((*p)++)->str); }

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