# include<iostream.h>
int main()
{
int ar[2]={8,2}; // (1)

int var1=66; //(2)

int var2=111; //(3)

int* ptarray[5]; //(4)

ptarray[0]= ar; //(5)

ptarray[1]= &ar[1]; //Address of second element of array ar // (5)

ptarray[2]= &var1; //(6)

ptarray[3]= &var2; //(7)

ptarray[4]= &ar[0]; //(8)

// To keep code small I use a loop //(9)

for(int i=0;i<5;i++)

cout<<*(ptarray[i])<<endl; // (10)
}


CASE1:

OUTPUT 1: 8
2
6
11
8

CASE 2: IF i replace *(ptarray[i]) with &(ptarray[i]) in LINE

NO 10 IN COUT << statement then output is

OUTPUT 2 : 0x 8fb8ffe4
0x 8fb8ffe6
0x 8fb8ffe8
0x 8fb8ffea
0x 8fb8ffec

CASE 3: IF I REPLACE *(ptarray[i]) with (ptarray[i]) in LINE
NO 10 THE OUT PUT IS
OUTPUT3:
0x8fb8fff2
0x8fb8fff4
0x8fb8fff0
0x8fb8fffe
0x8fb8fff2
question :
what is the difference between &ptarray[i] &
ptarray[i] due to that the address location of output 2 & output 3 is changing .

thanx in advaNCE
virendra

You should go throught pointer tutorial.
OK *(ptarray[i]) is dereferenced pointer of ptarray[i]. Here is the deal.
ptarray[0] value is address of ar. When you dereference it you will have the first element of the ar array which is 8. It is very similar to ptarray[4] becouse it's value is the address of the first element of ar array. Thats why *(ptarray[0]) and *(ptarray[4]) is equal as same as ptarray[0] and ptarray[4]. &ptarray[i] is the addreses of ptrarray[i].
So int* ptarray[5]; is array of pointers.

You defined an array of pointers to int named ptarray. Array's elements are pointers. So ptarray[i] represent array element which is actually address of some variable elswhere. For example, if address of variable "var" is placed to ptrarray as i+1 element, ptarray[i] referes to address of variable "var". On the other hand, &ptarray[i] is memory address of ptarray[i]. Arrays are also placed in memory and each array's elements has it's own address. So, &ptarray[i] represents memory address of array's element, and it's value (ptarray[i]) represents memory address of variable "var" because ptarray is an array of pointers. Remember, pointer is nothing more than variable that store memory address of some other variable.
I hope it helps.

ok ,so ptarray[i] the is the pointer array which is holidng the address of some variable and &(ptarray[i]) is the derefrence i.e. it holds the address of ptarray[i] location i.e.pointer to pointer
Is I am right if not please clear my doubt
and so, why the last digit of output 2 & 3 changes i.e.
For integer the adress changes with two bytes as in output 2 ok it is underastandable but in output 3 why the last digit of output address differ with more than two bytes i.e. 2 ,4 ,0,e(last digit of out put 3)
really thanks for these two reply

virendra
(intermediate programmer)

I assume your compiler is rather old.
Look at this code (modified loop)

And this is my output:

0x22ff30 0x22ff58
0x22ff34 0x22ff5c
0x22ff38 0x22ff54
0x22ff3c 0x22ff50
0x22ff40 0x22ff58

This makes sense. First column represents addresses of array elements. Since arrays are continuos in memory (on most compilers I work with, althought I'm not sure if this is guaranteed by the Standard) and on most 32 bit machines size(void*) is 4 bytes (this is also not guaranteed) first column rows are evenly spaced by 4 bytes.
Second column is a different story. First two rows represents addresses of array elements and difference is 4 byte. Also, last element of array hold address of first element of "ar" (same as ptarray[0]), so that's ok.
Other elements hold addresses of variables var1 and var2. There is no guarantee they will be evenly spaced, and probably in most cases they won't.
Is this clear to you?