Definite Iterations

[cheshirecat]

I want to have a program that finds the sum of a group of numbers. I would like for it to ask how many numbers total there will be, and then ask for the input.

So output like -


How many numbers do you want to input? 4

Input 1:1
Input 2:2.5
Input 3:10
Input 4:3

Sum = 16.5
Average = 4.13


-----
I read that you can use a definite iteration which takes the amount of times specified, and executes the statement that amount of times. So in this case, 4 would be the times specified. I don't know what code I would use to get that to happen though.

[pogrady]

its called a "for" loop:

for(int i = 0; i = /*some qualifying value*/; i++)
{


}

If you wanted to loop 4 times, the qualifying value would be

i < 4;

[vlad from moscow]

#include <iostream>

int main()
{
   unsigned int n;

   std::cout << "How many numbers do you want to input: ";
   std::cin >> n;

   double sum = 0.0;

   for ( unsigned int i = 0; i < 4; i++ )
   {
      double value = 0.0;
      std::cout << "Inpun #1: ";
      std::cin >> value;
      sum += value;
   }

   std::cout << "\nSum = " << sum << std::endl;
   std::cout << "Average = " << sum / n << std::endl;

   return 0;
}
 

[cheshirecat]

Alright, I took what you did, and rewrote it like this -

#include "stdafx.h"
#include <iostream>
using namespace std;

int _tmain(int argc, _TCHAR* argv[])
{
   int n;
   int x;

   cout << "How many numbers you want to input? ";
   cin >> n;
  
   double sum = 0.0;

   for (int i = 0; i < n; i++ )
   {
      double x = 0.0;
	  cout << "Input # "; cout << i+1; cout << "-->"; 
      cin >> x;
      sum += x;
   }

   cout << "\nSum = " << sum << endl;
   cout << "Average = " << sum / n << endl;

   return 0;
}

[cheshirecat]

Is there a way that I could do it so that so that I could just have it not ask for how many numbers, but just keep asking for a number? And then if they enter 0 it stops asking?

so example output would be:
Please enter a number:5
Please enter the next number: 21
Please enter the next number: 0

You have entered 0.

[cheshirecat]

Also, for the code I wrote above, is there a way to always round to the nearest 10th or 100th?

[cheshirecat]

Actually, I figured out the rounding to nearest 100th -

// project2.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include <iostream>
using namespace std;
#include <iomanip>

int _tmain(int argc, _TCHAR* argv[])
{
   int n;
   int x;

   cout << "How many numbers you want to input? ";
   cin >> n;

   double sum = 0.0;

   for ( unsigned int i = 0; i < n; i++ )
   {
      double x = 0.0;
          cout << "Input # " << i+1 << "-->";
      cin >> x;
      sum += x;
   }

   cout << setprecision(2) << fixed << "\nSum = " << sum << endl;
   cout << setprecision(2) << fixed << "Average = " << sum / n << endl;

   return 0;
}