Raising a number to a power in Java

Question

Here is my code. For some reason my BMI is not calculated correctly. When I check the output on a calculator for this : (10/((10/100)^2))) I get 1000, but in my program, I get 5. I'm not sure what I am doing wrong. Here is my code:

import javax.swing.*;

public class BMI {
    public static void main(String args[]) {
        int height;
        int weight;
        String getweight;
        getweight = JOptionPane.showInputDialog(null, "Please enter your weight in Kilograms");
        String getheight;
        getheight = JOptionPane.showInputDialog(null, "Please enter your height in Centimeters");
        weight = Integer.parseInt(getweight);
        height = Integer.parseInt(getheight);
        double bmi;
        bmi = (weight/((height/100)^2));
        JOptionPane.showMessageDialog(null, "Your BMI is: " + bmi);
    }
}

@Malfist As I read what you wrote I can see you waving your hand in front of his face like a jedi — John Vint, Jan 12 '12 at 22:20
Height/100 uses integer division, which might not be what you're looking for. — NomadMaker, Apr 25 '21 at 07:01

score 159 · Accepted Answer · edited Jan 27 '12 at 02:50

159

^ in java does not mean to raise to a power. It means XOR.

You can use java's Math.pow()

And you might want to consider using double instead of int—that is:

double height;
double weight;

Note that 199/100 evaluates to 1.

edited Jan 27 '12 at 02:50

Cody Gray - on strike

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answered Jan 12 '12 at 21:22

WuHoUnited

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I see, I knew it had to be something very simple that I was doing wrong. Thank you. – Jan 12 '12 at 21:23
In this case, it is probably OK, because accuracy doesn't matter for the seriously flawed BMI algorithm, sadly still used. When you do need decimal accuracy, better to use BigDecimal, because it doesn't suffer from inherent binary to decimal conversion errors of binary floating point numbers, because it uses decimal rather than binary scaling. – Infernoz Dec 17 '19 at 11:04

score 42 · Answer 2 · answered Jul 20 '16 at 18:02

42

we can use

Math.pow(2, 4);

this mean 2 to the power 4 (2^4)

answer = 16

answered Jul 20 '16 at 18:02

Thamays

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actually, when you only want powers of 2, use bitshifting: `1<<4 = 16` – xeruf Oct 27 '17 at 21:47
I would only use bit-shifting if performance is critical. Otherwise, it just makes the code harder to read. – Paul Wintz Apr 06 '19 at 07:34

score 12 · Answer 3 · edited Aug 07 '21 at 00:17

12

^ is not the operator you want. You are looking for the pow method of java.lang.Math.

You can use Math.pow(value, power).

Example:

Math.pow(23, 5); // 23 to the fifth power

edited Aug 07 '21 at 00:17

Kevin Hooke

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answered Jan 12 '12 at 21:23

lhk

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score 9 · Answer 4 · answered Jan 12 '12 at 21:24

9

Your calculation is likely the culprit. Try using:

bmi = weight / Math.pow(height / 100.0, 2.0);

Because both height and 100 are integers, you were likely getting the wrong answer when dividing. However, 100.0 is a double. I suggest you make weight a double as well. Also, the ^ operator is not for powers. Use the Math.pow() method instead.

answered Jan 12 '12 at 21:24

Bernard

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Thank you. So now instead of giving me 1000, it gives me 999.9999 when I enter 10 and 10 as weight and height. It seems to be working much better now though. bmi = weight / Math.pow(height/100.00,2.00); – stytown Jan 12 '12 at 21:31
1

That depends on the precision of your `double` values. You can always round as the final step in the calculation if necessary. – Bernard Jan 12 '12 at 21:41

score 4 · Answer 5 · answered Nov 28 '16 at 12:25

Too late for the OP of course, but still... Rearranging the expression as:

int bmi = (10000 * weight) / (height * height)

Eliminates all the floating point, and converts a division by a constant to a multiplication, which should execute faster. Integer precision is probably adequate for this application, but if it is not then:

double bmi = (10000.0 * weight) / (height * height)

would still be an improvement.

score 3 · Answer 6 · edited Apr 26 '21 at 14:05

3

You should use below method-

Math.pow(double a, double b)

From (https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#pow-double-double-)

Returns the value of the first argument raised to the power of the second argument.

edited Apr 26 '21 at 14:05

Mr R

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answered Jan 10 '19 at 08:18

PulkitRajput

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score 2 · Answer 7 · edited Nov 14 '19 at 07:56

1) We usually do not use int data types to height, weight, distance, temperature etc.(variables which can have decimal points) Therefore height, weight should be double or float. but double is more accurate than float when you have more decimal points

2) And instead of ^, you can change that calculation as below using Math.pow()

bmi = (weight/(Math.pow(height/100, 2)));

3) Math.pow() method has below definition

Math.pow(double var_1, double var_2);

Example:

i) Math.pow(8, 2) is produced 64 (8 to the power 2)

ii) Math.pow(8.2, 2.1) is produced 82.986813689753 (8.2 to the power 2.1)

score 2 · Answer 8 · answered Jan 12 '12 at 22:26

2

int weight=10;
int height=10;
double bmi;
bmi = weight / Math.pow(height / 100.0, 2.0);
System.out.println("bmi"+(bmi));
double result = bmi * 100;
result = Math.round(result);
result = result / 100;
System.out.println("result"+result);

answered Jan 12 '12 at 22:26

preeti

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score 1 · Answer 9 · answered Apr 25 '21 at 05:18

I did the benchmarking with Math.pow(x,2) and x*x, the result is that Math.pow() is easily forty times slower than manually multiplying it by itself, so i don't recommend it for anything where a little bit of performance is required.

Here's the results:

proof_of_work: 19.284756867003345
time for Math.pow(x,2) in ns: 35143
proof_of_work: 19.284756867003345
time for x*x in ns: 884
manual calculation is 39 times faster

and here's the test-code

double r1 = 4.391441320;
long multiply_d1 = System.nanoTime();
double multiply_dr = Math.pow(r1,2);
long multiply_d2 = System.nanoTime();
System.out.println(("proof_of_work: ") + (multiply_dr));
System.out.println(("time for Math.pow(x,2) in ns: ") + (multiply_d2 - multiply_d1));
long multiply_t1 = System.nanoTime();
double multiply_tr = r1*r1;
long multiply_t2 = System.nanoTime();
System.out.println(("proof_of_work: ") + (multiply_tr));
System.out.println(("time for x*x in ns: ") + (multiply_t2 - multiply_t1));
System.out.println(("manual calculation is ") + ((multiply_d2 - multiply_d1) / (multiply_t2 - multiply_t1)) + (" times faster"));

score 0 · Answer 10 · answered Nov 18 '21 at 16:59

Most efficient solution is

public Float fastPow(Float number, Integer power) {
        if (power == 0) {
            return 1.0f;
        } else if (power % 2 == 1) {
            return fastPow(number, power - 1) * number;
        } else {
            return fastPow(number * number, power / 2);
        }
    }

Let A is our number and N our power. Then A^2^N = (A^2)^N. And A^N = (A^2)^N/2. The function above reflects this relationship.

Raising a number to a power in Java

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