Using for loops to create a bar graph - c++

Question

CAN ONLY USE LOOPS & IF/ELSE Statements
So for my assignment I'm given this: Write a program with for loops that asks the user to enter today’s sales for five stores. The program should then display a bar graph comparing each store’s sales. Create each bar in the bar graph by displaying a row of asterisks. Each asterisk should represent $100 of sales.
So it should like:
Store 1: *****
Store 2: ***
Store 3: *******
etc.
I have most of it written up and everything displays properly except for my asterisks. I can't seem to get them to display. Also, I've searched around and there are answers to this, but all of them include arrays and we aren't allowed to use them in this exercise. Here's my code so far:
#include <iostream>
#include <iomanip>
using namespace std;
int main() {
int sales = 0, sale = 0; //Identfy Variables
for (int i = 1; i <= 5; i++) {
cout << "Enter today's sales for Store " << i << ":" << endl;
cin >> sale;
sales += sale;
sales /= 100; //in a similar manner to +=, /= can be used for division.
}
cout << "SALES BAR CHART:" << endl;
cout << "Each * = $100" << endl;
for (int x = 1; x <= 5; x++) {
cout << "Store " << x << ": ";
for (int s = 0; s < sales; s++) cout << "*";
cout << endl;
}
return 0;
}

Answer 1

Try this.
#include <iostream>
#include <string>
using namespace std;
int main()
{
string star = "";
int sale;
for (int i = 1; i <= 3; i++)
{
cout << "Enter today's sales for Store " << i << ":" << endl;
cin >> sale;
star += "Store " + to_string(i) + string(":");
for(int j= 0; j<sale/100; j++) {
star += "*";
}
star += "\n";
}
cout<<star;
return 0;
}

Answer 2

The problem with your code is that you do not store the data for each store individually. You have one variable called sales, which stores the sales of only the last store the end-user entered. All numbers the user enters before that get written over.
Since you have five stores, sales needs to be an array of five items:
int sales[5], total = 0;
Change your for loop to iterate 0 through 4, inclusive. Print (i+1) to end-users to stay one-based. Read sales numbers into sales[i] array element, then use it in the second loop to print your bar graph:
for (int i = 0 ; i != 5 ; i++) {
cout << "Enter today's sales for Store " << (i+1) << ":" << endl;
cin >> sales[i];
sales[i] /= 100;
total += sales[i];
}
...
// Now print your bar graph using a loop on i, and sales[i]

Answer 3

Your for loop is missing a definition of s:
for (int s; s < sales; s++)
should be:
for (int s = 0; s < sales; s++)
Without this explicit initialization, a variable s is declared, and memory is allocated for it, but its value remains whatever was left in memory from the last time that address was used. It's likely that this left over data was equivalent to an integer larger than sales, and thus your for loop wouldn't run, not even once.
As #dasblinkenlight mentioned, you have a singular variable that stores the "sales" of every store. Each store needs a separate variable, since each store is going to sell a different amount.

Answer 4

You already have the answers, I just want to add two options that would make printing the *'s easier:
using std::fill_n:
std::fill_n(std::ostream_iterator<char>(cout), sales, '*');
cout << endl;
using std::string's constructor overload (2):
cout << std::string(sales, '*') << endl;