HELP! im looking for different ways .

Im looking for ways to use arrays using pointers.
an ex is:
#include<iostream.h>
#include<conio.h>
void main ()
{
int sales[7]= {1,2,3,4,5,6,7}
int*p=&sales[0];
int x;
//-----------------------------
for(x=0; x<7; ++x)
cout<<"$"<<sales[x]<<" ";
cout<<endl;
//-----------------------------
for(x=0; x<7; ++x)
cout<<"$"<<p[x]<<" ";
cout<<endl;
//-----------------------------
for(x=0; x<7; ++x)
cout<<"$"<<*(sales+x<<" ";
cout<<endl;
//-----------------------------
for(x=0; x<7; ++x)
cout<<"$"<<*(p+x)<< " ";
cout<<endl;
getch();
}

\\ all of the above give the same result
\\$1 $2 $3 $4 $5 $6 $7

any more ways???????
at least I need 10 more ways

any feed back is greatly accepted

Quote:

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Originally posted by moenia
Im looking for ways to use arrays using pointers.
an ex is:
int num[5]={2,6,9,2,2};
int*ptr;
ptr=num;
ptr=&num[0]



any feed back is greatly accepted

---

congrats, you just ruined your pointer.

Code:

---

int num[5] = {2, 6, 9, 2, 2};
int *ptr = num[1];
cout<<ptr;

---

output :

Code:

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6

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using & means its appointed to a reference so:

&num = numb;

would not make &num = numb, it would just replace it when it comes up in the program so:

cout<<&num;

would display numb and not &num

cout<<num;

would display the address of the variable numb
ex:
output : Ax00A4b00

(maybe its not a real address i dunno but it looks kinda like that)

dont try modifying addresses cause u could mess up your computer -_-'