CODEWITHSUNDEEP
c++doubleaverage
hennessy
hennessy

Reputation: 331

C++ Cant get to complete the average calculation with double values

What this code does now: I give int values and it calculates the average between them. What Ive spent hours on trying to get it do: Ive tried making it so that it would calculate the average between double values. Ive tried everything but it always fails or goes into an infinite loop or will not compile.

Question: So how should I modify my code to make it work with double values/numbers?

#include <stdio.h>


void main()
{
int Tau[10]={0,0,0,0,0,0,0,0,0,0};
int r, i = 0;
int m = 0;
int huku = 0;


do{

printf("Enter numbers: ");
scanf_s("%d", &i);
Tau[m]+=i;
huku++;


}while(i != 0);

r = (Tau[m]/(huku-1));


printf("The average of your numbers is; %d\n", r);
}

Upvotes: 0

Views: 3361

Answers (6)

DarioM
DarioM

Reputation: 13

// Question is tagged for C++, but the code is in C.
// I will change your code a bit, because you had quite a few mistakes.
#include <stdio.h>

void main ()
{
    int sum = 0; // sum of all numbers you entered, to find average you only need total sum and number of entries
    int numOfEntries; // number of entries (numbers taken from input)
    int inputNum; // variable where you will write numbers from input one by one
    double average; // Not really needed, but it can help to simplify the problem to you.

    printf("Enter numbers: ");
    do
    {
        scanf_s("%d", &inputNum);
        sum += inputNum;
        numOfEntries++;
    } while (inputNum != 0); // I understand you read numbers until you read value 0.
    // int / int will give you rounded number, not the true average, so we need to convert one of the operands to a real number, in this case double
    // double / int or int / double will give you a real number as result, which will have true average value, and that is why I converted sum to a real number
    if (numOfEntries != 0)
        average = (double)sum / numOfEntries;
    else
        average = 0;
    printf("The average of your numbers is; %f\n", average); // Here I did it again - print double instead of int to get true value.
}

It will be even easier to change this:

....
double sum = 0;
...
average = sum / numOfEntries; // Here sum is already double, not int, so you don't need to change it manually.
...

Now, if you want to make it work for double, the only difference will be:

    double sum = 0;
    double inputNum;
    scanf_s("%lf", &inputNum);
    average = sum / numOfEntries;

So, to round up the story - you have variable to input a number from keyboard, a variable which holds sum of all entered numbers so far, a variable which counts how many numbers you entered from keyboard. You input numbers until you enter 0 as value, then the program will exit the loop. Formula for average number is sum of all divided by number of numbers. With integers you have to add conversion to a real number or otherways you won't get accurate result.

I hope I didn't confuse you. :D

Upvotes: 0

digit plumber
digit plumber

Reputation: 1172

double r = 0;
int i = 0;
r = ((double)Tau[m]/((double)huku-1));
printf("The average of your numbers is; %f\n", r);

Upvotes: 0

Pete Becker
Pete Becker

Reputation: 76275

The simplest changes only involve four lines:

double Tau[10] = {0,0,0,0,0,0,0,0,0,0};
double r, i = 0;
scanf_s("%f", &i);
printf("The average of your numbers is; %f\n", r);

Note that this doesn't address the coding issues; all it does is change the code to read and work with double instead of int.

Upvotes: 1

Florin Petriuc
Florin Petriuc

Reputation: 1176

  1. I don't see why you need Tau to be a vector in your case. An int would suffice.
  2. huku-1 -> That is wrong. It should be just huku.
  3. You need to cast Tau and huku to double when you divide. It won't hurt to check that huku is != 0 also.
  4. m is useless. Just delete it.
  5. r shouldn't be an int if you wish to store Tau/huku in it.
  6. in printf replace %d with %lf

Upvotes: 1

mathematician1975
mathematician1975

Reputation: 21351

You have some issues in your code but basically, integer division will not give you doubles. The result of an integer divided by an integer is another integer, not a double. If you want doubles you need to cast either the numerator or denominator to a double and store the result in a double.

Upvotes: 3

Steve Wellens
Steve Wellens

Reputation: 20620

Welcome to Numerical Analysis 1001.

Integer math:

2 / 3 = 0;
4 / 2 = 2;
5 / 2 = 2;

Integers don't do fractions.

Upvotes: 1

Related Questions