CODEWITHSUNDEEP
c
LookingWest
LookingWest

Reputation: 69

Square root not diving properly

I've created a program which takes an integer x input, then loops until x is met while also taking other integer inputs. I then do various calculations, and then find a square root of a certain value. When I divide by square root however I get a 0 when I know I should be getting a different value as the maths doesn't add up.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>  

int main(void) {

    int multiply1, multiply2, add, squareRoot;
    int i;
    int n;
    int x;
    int s;
    double divide, test = 0;

    scanf("%d", &x);

    for (s = 0; s < x; s++) {
        scanf("%d %d", &i ,&n);
    }

    multiply1 = i * i;
    multiply2 = n * n;

    add = multiply1 + multiply2;

    squareRoot = sqrt(add);
    printf("%d", i);
    test = (i / squareRoot);

    printf("Multiplication = %d\n", multiply1);
    printf("Multiplication = %d\n", multiply2);
    printf("Added together = %d\n", add);
    printf("square root = %d\n", squareRoot);
    printf("First output = %.3f\n", test);

    return 0;
}

Upvotes: 1

Views: 92

Answers (2)

Iharob Al Asimi
Iharob Al Asimi

Reputation: 53016

There are two things you can do,

  1. Without changing your program, simply cast the i and squareRoot variables to double

    test = (double) i / (double) squareRoot;

  2. Change your program and make i and squareRoot a double.

I, would choose 2 because sqrt() returns a double and that might cause an integer overflow.

Upvotes: 1

mcrlc
mcrlc

Reputation: 137

You are dividing two integers so the actual division returns the result rounded down. You should instead cast to double and then divide.

test = ((double)i/squareRoot);

Upvotes: 2

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