Why do I keep getting 0 as an answer to my program even though it already has its values? (C++)

Question

The code line cout.width(10); cout << right << (freq) /(n)*100 <<"\t"; specifically is always getting 0 as a value. I'm trying to get the relative frequency of the table. Please help and explain, I don't understand why it keeps getting 0 as an output even though freq and n has its respected values.

#include <iostream>
#include <iomanip>
#include <math.h>

using namespace std;

int main()
{
    int i, j, nci, n, freq = 0;
    float temp, min, max, range, interval, cibase;

    cout << "Enter the Sample Size: ";
    cin >> n;

    float x[n];
    cout << "Enter the Data: ";
    for(i = 0; i < n; ++i){
        cin >> x[i];
    }

    cout << endl;
    cout << "Number of Class Interval (6 - 20): ";
    cin  >> nci;

    if(nci >= 5 && nci <= 20){
        for(i = 0; i < n; ++i){
            for(j = 1 + i; j < n; ++j){
                if(x[i] > x[j]){
                    temp = x[i];
                    x[i] = x[j];
                    x[j] = temp;
                }
            }
        }

        cout << endl;

        min = x[0];
        max = x[n - 1];
        range = max - min;
        interval = ceil(range / nci);

        cout << " Class Interval \t";
        cout << " Freq \t";
        cout << "Xi \t" ;
        cout << " Class Boundary \t";
        cout << " Class Mark " << endl;

        cibase = min;
        while(cibase <= max){
            cout.width(5); cout << right << cibase << " - ";
            cout.width(4); cout << right << (cibase - 1) + interval << "\t\t";
            freq = 0;
            for(i = 0; i < n; ++i){
                if(x[i] >= cibase && x[i] <= ((cibase - 1) + interval)){
                    ++freq;
                }
            }
           
            
            
            cout.width(4); cout << right << freq << "\t";
            cout.width(7); cout << right <<(cibase + interval/2) -.5 <<"\t";
            cout.width(6); cout << right << cibase - 0.5 << " - ";
            cout.width(5); cout << right << (cibase - 0.5) + interval << "\t\t";
            cout.width(9); cout << right << (cibase + ((cibase - 1) + interval)) / 2 << "\t";
            cout.width(10); cout << right << (freq) /(n)*100 <<"\t";
            cout <<  "\n" << endl;
            cibase = cibase + interval;
        }

        cout.width(6); cout << "Minimum: " << setprecision(5) << right << min << endl;
        cout.width(6); cout << "Maximum: " << setprecision(5) << right << max << endl;
        cout.width(6); cout << "Range: " << setprecision(5) << right << range << endl;
        cout.width(6); cout << "Interval: " << setprecision(5) << right << interval << endl;
    }
    else{
        cout << "ERROR: Choose Between 7 and 20" << endl;
        cout << "Try Again!" << endl;
    }

    return 0;
}

You cant do `float x[n];` in C++, use `std::vector x(n)` instead. — D-RAJ, Jan 26 '21 at 18:42

ΦXocę 웃 Пepeúpa ツ · Answer 1 · 2021-01-26T18:48:07.990

1

short answer: do replace 100 with 100.0

(freq) /(n)*100.0

long answer:

dividing integer data types produce an integer as data type too, so if you do A/B and B>A and both B and A are integers, the result is 0

edited Jan 26 '21 at 18:48

answered Jan 26 '21 at 18:42

ΦXocę 웃 Пepeúpa ツ

47,427
17
69
97

It still results in 0 even if I change it to 100.0 . Is there any other way? – Miks Jan 26 '21 at 18:50
nope... it must be 0.00 if not then you are doing something wrong... – ΦXocę 웃 Пepeúpa ツ Jan 26 '21 at 18:52
1

Multiplication and division group left to right, so that expression is equivalent to `(freq/n) * 100.0`. The division still divides two integers. Write it as `100.0 * freq / n` to take advantage of the grouping rules. – Pete Becker Jan 26 '21 at 19:57
Thank you Pete! Didn't know about that. – Miks Jan 27 '21 at 04:00

Why do I keep getting 0 as an answer to my program even though it already has its values? (C++)

1 Answers1