[Steve18]

I'm very new to programming and ive been trying to figure out this problem for a few days and still have no luck, I cant figure out how to get the average to ignore an out of range number. Also the range is the max- the min which i cant get to work either please help.

#include <iostream>
using namespace std;

int main()
{
double x;
double sum = 0.0;
int number = 0;
double average;
double max = 0;
double min = 0;

while (cin>>x)

{
if ((x < 0) || (x > 100))
{
cin.ignore(3,'\n');

cout << "Out of range ; ignored." << endl;
}

if (x > max)
max = x;

if (min > x)
x = min;

sum += x;
number++;
}
if (number > 0)
{
average = sum / number;
cout<<"The average is "<< average;
cout<<" The range is "<< max - min;
cout << endl;
}
return 0;
}

[asha]

What's the objective of the program??

[oogabooga]

You're starting min at 0, so nothing will be less than it. You need to start it at the maximum (100).

You need to add "continue;" after the out of range check to skip the rest of the while loop:

if ((x < 0) || (x > 100))
{
    cin.ignore(99,'\n');
    cout << "Out of range ; ignored." << endl;
    continue;
}

The line x = min needs to be min = x.