### count with array

Hello people, now I'm learning array.
I now know how to count integers with while loop
but I'm not sure how to count the integers with array.
so the question is:
1. program should keep reading integers as long as the integers are within [0,9999]
2. when user typed the integer not between 0 to 9999, the program print out the numbers of integers that were typed.

Sample
3
3
3
9999
9999
-1
You entered 3 3 times.
You entered 9999 2 times.

Can anyone help me with this, I think it is really simple, but I'm stuck from the beggining
 ``12345678910111213`` ``````#include using namespace std; int main () { int numbers[10000]; int i; int n; for (i = 0; i >= 0; i = i+1){ cin >> numbers[i]; } for ( cout << "You entered " << i << " " << n << " times.\n"; return 0; }``````
I think you should use two arrays. One that stores the numbers from 0 - 9999 in one array and the second that stores the numbers of times a certain number from the array is chosen.
I check the first array, and it didn't end when I typed -1 or 10000.
I think there is a problem with the first array, can you tell me what is wrong?
`for (i = 0; i >= 0; i = i+1)`

If i starts at 0 and continues to be incremented, your condition in the for loop will always be true. It's an infinite loop.
@wildblue
than it's gotta be
 `` `` ``for (i = 0; i <=9999; i = i+1)``

but the problem is that i cannot go below 0
I'm thinking that the user input should be in a while loop - - - you don't know how many times they are going to enter numbers right?

Running a for loop to initialize an array of all possible values could be done with the for loop you have but without the `cin` - that would mandate that someone enter 10000 values by hand. I don't think you need to do that though.

I think you can just do this with one array with 10000 elements (initialized to 0), and then increment the value of the element corresponding to the number entered by the user, e.g. if they enter 2000, array[2000] gets incremented by 1.
@wildblue
so it's
 ``123456789101112131415161718192021`` ``````#include using namespace std; int main() { int i=-1; int x; int numbers[10000]; for(x=0; x<10000; x=x+1) numbers[x] = 0; do{ if(i != -1) numbers[i]=numbers[i]+1; cin >> i; } while(i>-1 && i<10000);{ for(x=0; x<10000; x=x+1) { if(numbers[x]) cout << "You entered " << x << " " << numbers[x] << " times.\n"; } } return 0; }``````

my friend helped me doing it,he did the do part, but I didn't learn "do" yet
how can i remove "do"?
Last edited on
You can initialize an array to all 0 values without having to go through a for loop.

`int numbers[10000]={};`

just using a while, not do while...

ask user to enter value within range (i)
while (number is within accepted range)
---array[i] increment by one
You can't just take out the `do` in its current location. The while loop syntax starts with a while statement.