int numbers[10];for(int i=0; i<10; ++i){   cin >> numbers;} 

the ''int numbers[10]'' bit creates an array of 10 integers.
you access an element in the array by putting the number of the element you want into square brackets ''[]''... eg numbers[3] will give you the 4th integer in the array (since the indexes start at 0)

be careful though - C++ does no bounds checking, so ''numbers[54] = 3'' will compile fine, but will probalby crash when you run it.


simple as that.

that last post was me but since i forgot to give my details, i cant edit it... so heres a brand new post

#include int numbers[10];for(int x=0; x<10; ++x){   cin >> numbers[x];}//small usage examplesnumbers[0] = 2;  //finenumbers[53] = 3; //will compile ok, but will crash at run timeif(numbers[3] == 3)   cout << "the 4th element stores the number 3" << endl;  

Edited by - POINT on June 16, 2000 5:22:18 PM

Edited by - POINT on June 16, 2000 5:23:24 PM

I just made that post up as I went along.....
I had no intention for someone to figure it out.....
Did you really go through the trouble of seeing what happens ?
I guess you could just put it inside of a main loop, compile it and see what happens.


"Now go away or I shall taunt you a second time"
- Monty Python and the Holy Grail
themGames Productions

Its pretty simple if you really look at it. By the way, its an infinite loop.