We Keep Coding

How to retrieve decimal value from Integer?

I'm trying to calculate the average based on user input. Now I've got the basics working, but I am only getting whole numbers as result. I want decimal numbers like 5.5
Any explanation is welcome!
final TextView averageView = findViewById(R.id.averageView);
final String averageText = getString(R.string.average);
final Button calculateButton = findViewById(R.id.calculateAverageButton);
calaculateButton.setOnClickListener(new View.onClickListener() {
#SupressLint("SetTextI18n")
public void onClick(View v) {
int grade[] = {Integer.parseInt(((EditText) findViewById(R.id.grade1)).getText().toString());
int weight[] = {Integer.parseInt(((EditText) findViewById(R.id.weight1)).getText().toString());
int weightTotal = weight[0];
int sum = grade[0] * weight[0]
int average = sum / weightTotal
averageView.setText(averageText + Integer.toString(average));
EDIT:
I have experimented with a answer and got a solution for my problem.
final TextView averageView = findViewById(R.id.averageView);
final String averageText = getString(R.string.average);
final Button calculateButton = findViewById(R.id.calculateAverageButton);
calaculateButton.setOnClickListener(new View.onClickListener() {
#SupressLint("SetTextI18n")
public void onClick(View v) {
double grade[] = {Double.parseDouble(((EditText) findViewById(R.id.grade1)).getText().toString());
double weight[] = {Double.parseDouble(((EditText) findViewById(R.id.weight1)).getText().toString());
double weightTotal = weight[0];
double sum = grade[0] * weight[0]
double average = sum / weightTotal
averageView.setText(averageText + " " + Double.toString(average));

Ok You can do it like this
final TextView averageView = findViewById(R.id.averageView);
final String averageText = getString(R.string.average);
final Button calculateButton = findViewById(R.id.calculateAverageButton);
calaculateButton.setOnClickListener(new View.onClickListener() {
#SupressLint("SetTextI18n")
public void onClick(View v) {
int grade = {Integer.parseInt(((EditText)
findViewById(R.id.grade1)).getText().toString());
int weight = {Integer.parseInt(((EditText)
findViewById(R.id.weight1)).getText().toString());
double sum = grade * weight
double average = sum / weight
averageView.setText(averageText +" "+ average);

The division operator / means integer division if there is an integer
on both sides of it. If one or two sides has a floating point number,
then it means floating point division. The result of integer division
is always an integer. Integer division determines how many times one
integer goes into another. The remainder after integer division is
simply dropped, no matter how big it is.
Change the type of variable average to float and typecast one of the sum or weight to float:
float average = (float)sum / weightTotal;
Also change Integer.toString(average) to Float.toString(average)

You are getting integers because you are only using integers !!.
That's normal, If you like average as decimal you have to declare it as Float or Double instead of int.
Note that even if you use such casting (Float.parseFloat....) it wont work because the variable average can only hold integers.

float x = (float) y / z
text.setText("your avarage is : " + x );
also you should use try/catch while get data from edittext for convert Integer. otherwise you can get crash

How to move the digit in a double to the end without converting to string in Java

I am trying to make my number move the first character to the end.
For example I would have my double d.
double d = 12345.6;
double result = 2345.61;
Currently I am removing the first character of d with:
d = d % (int) Math.pow(10, (int) Math.log10(d));
But I do not know how to store the character I am removing so I could put it at the end. I know I could just convert d into an array or a string, but I want to keep it as a double if at all possible.
I am getting my double from a nanosecond clock using Instant.now, so I can guarantee it starts as an 8 digit positive int, which I start by adding .0 to so I can make it a double. I know I can just use string (as I mentioned in the post), but I was wondering if there was a way to do it without conversions.
Thanks for helping!
(this is my first post I apologize if it is bad)

This is a pretty tough problem, and this isn't a working answer but it is pretty close. The only trouble I ran into was java adding decimals to the end of my doubles because it can't represent a number very well. My solution might get you on the right path. The appendChar was messing up because it was adding digits to the end of the double, other than that problem this would have worked.
public static void main(String[] args) {
double test = 4234.1211;
boolean hasDecimals = test % 1 > 0;
double[] leadChar = getLeadDigitAndMulti(test);
double appendedValue = appendLeadChar(test, (int) leadChar[1], hasDecimals);
}
public static double[] getLeadDigitAndMulti(double value) {
double[] result = new double[2];
int multiplier = 10;
int currentMultiplier = 1;
while (value > currentMultiplier) {
currentMultiplier *= multiplier;
}
currentMultiplier /= multiplier;
double trail = value % (currentMultiplier);
result[0] = currentMultiplier;
double lead = value - trail;
lead /= currentMultiplier;
result[1] = lead;
return result;
}
public static double appendLeadChar(double value, int leadChar, boolean hasDecimals) {
if (!hasDecimals) {
return (value * 10) + leadChar;
}
int multiplier = 10;
double currentMultiplier = 1;
while (value > 0) {
value %= currentMultiplier;
currentMultiplier = currentMultiplier / multiplier;
}
return 0;
}

I would start with at least figuring out how many digits long your double is. I'm not too sure of how to make it work but I saw another question that would answer it. Here is the link to that thread:
Number of decimal digits in a double
And the closest code on there that I could find to find the length would be:
double d= 234.12413;
String text = Double.toString(Math.abs(d));
int integerPlaces = text.indexOf('.');
int decimalPlaces = text.length() - integerPlaces - 1;
NOTE THIS SEGMENT OF CODE IS NOT MINE, credit goes to Peter Lawrey who originally answered this in the linked thread.
Then using code or modified code similar to above ^ assuming you would find how to find the length of the double or the total number of digits of your double, you would create a new int or double variable that would store the double and round it using BigDecimal to the highest place value. The place that the BigDecimal will round to will be set using a 10*pow(x) where x is the number of places AFTER the decimal.
Sorry I am unable to provide code on how to make this actually run because I am pretty new to java. Here is what I mean in more detailed pseudocode:
double d = 12345.6; //the double you are evaluating
double dNoDecimal; //will be rounded to have no decimal places in order to calculate place of the 1st digit
double dRoundedFirstDigit; //stores the rounded number with only the 1st digit
double firstDigit; //stores first digit
dNoDecimal = d; //transfers value
dNoDecimal = [use BigDecimal to round so no decimal places will be left]
//dNoDecimal would be equal to 12345 right now
[use above code to find total amount of place values]
[knowing place values use BigDecimal to round and isolate 1st digit of double]
dRoundedFirstDigit = dNoDecimal; //transfers value
dRoundedFirstDigit = [round using BigDecimal to leave only 1st digit]
firstDigit = [dRoundedFirstDigit cut down to 1st digit value using length found from before, again an equation will be needed]
//Now use reverse process
//do the same thing as above, but find the number of decimals by rounding down to leave only decimals (leaving only .6)
//find the length
//use length to calculate what to add to the original value to move the digit to the end (in this case would be d += firstDigit*Math.pow(10,-2))
I apologize if I just made everything seem more confusing, but I tried my best, hope this helps.
Also, doing this without an array or string is really difficult... may I ask why you need to do it without one?

Get a numer decimal part as Integer using only math

Edit: This has to do with how computers handle floating point operations, a fact that every programmer faces once in a lifetime. I didn't understand this correctly when I asked the question.
I know the simplest way to start dealing with this would be:
val floatNumber: Float = 123.456f
val decimalPart = floatNumber - floatNumber.toInt() //This would be 0.456 (I don't care about precision as this is not the main objective of my question)
Now in a real world with a pen and a piece of paper, if I want to "convert" the decimal part 0.456 to integer, I just need to multiply 0.456 * 1000, and I get the desired result, which is 456 (an integer number).
Many proposed solutions suggest splitting the number as string and extracting the decimal part this way, but I need the solution to be obtained mathematically, not using strings.
Given a number, with an unknown number of decimals (convert to string and counting chars after . or , is not acceptable), I need to "extract" it's decimal part as an integer using only math.
Read questions like this with no luck:
How to get the decimal part of a float?
How to extract fractional digits of double/BigDecimal
If someone knows a kotlin language solution, it would be great. I will post this question also on the math platform just in case.
How do I get whole and fractional parts from double in JSP/Java?
Update:
Is there a "mathematical" way to "calculate" how many decimals a number has? (It is obvious when you convert to string and count the chars, but I need to avoid using strings) It would be great cause calculating: decimal (0.456) * 10 * number of decimals(3) will produce the desired result.
Update 2
This is not my use-case, but I guess it will clarify the idea:
Suppose you want to calculate a constant(such as PI), and want to return an integer with at most 50 digits of the decimal part of the constant. The constant doesn't have to be necessarily infinite (can be for example 0.5, in which case "5" will be returned)

I would just multiply the fractional number by 10 (or move the decimal point to the right) until it has no fractional part left:
public static long fractionalDigitsLong(BigDecimal value) {
BigDecimal fractional = value.remainder(BigDecimal.ONE);
long digits;
do {
fractional = fractional.movePointRight(1); // or multiply(BigDecimal.TEN)
digits = fractional.longValue();
} while (fractional.compareTo(BigDecimal.valueOf(digits)) != 0);
return digits;
}
Note 1: using BigDecimal to avoid floating point precision problems
Note 2: using compareTo since equals also compares the scale ("0.0" not equals "0.00")
(sure the BigDecimal already knows the size of the fractional part, just the value returned by scale())
Complement:
If using BigDecimal the whole problem can be compressed to:
public static BigInteger fractionalDigits(BigDecimal value) {
return value.remainder(BigDecimal.ONE).stripTrailingZeros().unscaledValue();
}
stripping zeros can be suppressed if desired

I am not sure if it counts against you on this specific problem if you use some String converters with a method(). That is one way to get the proper answer. I know that you stated you couldn't use String, but would you be able to use Strings within a Custom made method? That could get you the answer that you need with precision. Here is the class that could help us convert the number:
class NumConvert{
String theNum;
public NumConvert(String theNum) {
this.theNum = theNum;
}
public int convert() {
String a = String.valueOf(theNum);
String[] b = a.split("\\.");
String b2 = b[1];
int zeros = b2.length();
String num = "1";
for(int x = 0; x < zeros; x++) {
num += "0";
}
float c = Float.parseFloat(theNum);
int multiply = Integer.parseInt(num);
float answer = c - (int)c;
int integerForm = (int)(answer * multiply);
return integerForm;
}
}
Then within your main class:
public class ChapterOneBasics {
public static void main(String[] args) throws java.io.IOException{
NumConvert n = new NumConvert("123.456");
NumConvert q = new NumConvert("123.45600128");
System.out.println(q.convert());
System.out.println(n.convert());
}
}
output:
45600128
456

Float or Double are imprecise, just an approximation - without precision. Hence 12.345 is somewhere between 12.3449... and 12.3450... .
This means that 12.340 cannot be distinghuished from 12.34. The "decimal part" would be 34 divided by 100.
Also 12.01 would have a "decimal part" 1 divided by 100, and too 12.1 would have 1 divided by 10.
So a complete algorith would be (using java):
int[] decimalsAndDivider(double x) {
int decimalPart = 0;
int divider = 1;
final double EPS = 0.001;
for (;;) {
double error = x - (int)x;
if (-EPS < error && error < EPS) {
break;
}
x *= 10;
decimalPart = 10 * decimalPart + ((int)(x + EPS) % 10);
divider *= 10;
}
return new int[] { decimalPart, divider };
}

I posted the below solution yesterday after testing it for a while, and later found that it does not always work due to problems regarding precision of floats, doubles and bigdecimals. My conclusion is that this problem is unsolvable if you want infinite precision:
So I re-post the code just for reference:
fun getDecimalCounter(d: Double): Int {
var temp = d
var tempInt = Math.floor(d)
var counter = 0
while ((temp - tempInt) > 0.0 ) {
temp *= 10
tempInt = Math.floor(temp)
counter++
}
return counter
}
fun main(args: Array <String> ) {
var d = 3.14159
if (d < 0) d = -d
val decimalCounter = getDecimalCounter(d)
val decimalPart = (d - Math.floor(d))
var decimalPartInt = Math.round(decimalPart * 10.0.pow(decimalCounter))
while (decimalPartInt % 10 == 0L) {
decimalPartInt /= 10
}
println(decimalPartInt)
}
I dropped floats because of lesser precision and used doubles.
The final rounding is also necessary due to precision.

Problems Generating A Math.random Number, Either 0 or 1

I want a random number, either 0 or 1 and then that will be returned to main() as in my code below.
import java.util.Scanner;
public class Exercise8Lab7 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int numFlips = 0;
int heads = 0;
int tails = 0;
String answer;
System.out.print("Please Enter The Number Of Coin Tosses You Want: ");
numFlips = input.nextInt();
for(int x = 1;x <= numFlips; x++){
if(coinToss() == 1){
answer = "Tails";
tails++;
}
else{
answer = "Heads";
heads++;
}
System.out.print("\nCoin Toss " + x + ": " + answer);
}
System.out.println("\n\n====== Overall Results ======" +
"\nPercentage Of Heads: " + (heads/numFlips)*100 + "\nPercentage Of Tails: " + (tails/numFlips)*100);
}
public static int coinToss(){
double rAsFloat = 1 * (2 + Math.random( ) );
int r = (int)rAsFloat;
return r;
}
}
Many solutions had been suggested to use the util.Random option which I have done and works perfectly but I want to sort out why I can't get this to work. Obviously I want the number to be an int myself so I convert it to an int after the random number has been generated. But no matter what I add or multiply the Math.random() by, it will always all either be Heads or all either be Tails. Never mixed.

Try this) It will generate number 0 or 1
Math.round( Math.random() ) ;

You could use boolean values of 0 or 1 based on value of Math.random() as a double between 0.0 and 1.0 and make the random generator much simpler. And you can get rid completely of the coinToss() method.
if(Math.random() < 0.5) {
answer = "Tails";
tails++;
}
Remove the coin toss method and replace the first conditional with the code above.
Math.random(); by itself will return a value between 0.0 and less than 1.0. If the value is in the lower half, [0.0, 0.5), then it has the same probability of being in the upper half, [0.5, 1.0). Therefore you can set any value in the lower half as true and upper as false.

Wierd that no one is using a modulo division for the random number.
This is the simplest implementation you can get:
Random rand = new Random();
int randomValue = rand.nextInt() % 2;

Math.round(Math.random()) will return either 0.0 and 1.0. Since both these values are well within the limits of int range they can be casted to int.
public static int coinToss(){
return (int)Math.round(Math.random());
}

(int)(Math.random()*2) also works fine in this case

its not working because of the integer math you are using, the call to 2+ Math.Random is pretty much always giving you a answer between 0.0 and 1.0.
so assuming that you recieve 0.25 as your result your maths is as follows
double d = 1* (2 + 0.25); // (result = 2
Then you are checking to see if your result == 1 ( which it never will. )
A better result would be to declare java.util.Random as a class variable and call random.nextBoolean() and simply perform your heads/tails calculation on that.
If you were to continue to use Math.random() and lets say
return Math.random() < 0.5
Your results would be ever so slightly skewed due to the fact that Math.random() cannot return 1.0, due to the fact that the java API specification states:
"Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0."

Math.random() returns a random float in the range [0.0,1.0)--that means the result can be anything from 0 up to but not including 1.0.
Your code
double rAsFloat = 1 * (2 + Math.random( ) );
will take this number in the [0.0,1.0) range; adding 2 to it gives you a number in the [2.0,3.0) range; multiplying it by 1 does nothing useful; then, when you truncate it to an integer, the result is always 2.
To get integers from this kind of random function, you need to figure out how many different integers you could return, then multiply your random number by that. If you want a "0 or 1" answer, your range is 2 different integers, so multiply Math.random() by 2:
double rAsFloat = 2 * Math.random();
This gives you a random number in the range [0.0,2.0), which can then be 0 or 1 when you truncate to an integer with (int). If, instead, you wanted something that returns 1 or 2, for example, you'd just add 1 to it:
double rAsFloat = 1 + 2 * Math.random();
I think you've already figured out that the Random class gives you what you want a lot more easily. I've decided to explain all this anyway, because someday you might work on a legacy system in some old language where you really do need to work with a [0.0,1.0) random value. (OK, maybe that's not too likely any more, but who knows.)

The problem can be translated to boolean generation as follow :
public static byte get0Or1 {
Random random = new Random();
boolean res= random.nextBoolean();
if(res)return 1;
else return 0;
}

Here it the easiest way I found without using java.util.Random.
Blockquote
Scanner input = new Scanner (System.in);
System.out.println("Please enter 0 for heads or 1 for tails");
int integer = input.nextInt();
input.close();
int random = (int) (Math.random() + 0.5);
if (random == integer) {
System.out.println("correct");
}
else {
System.out.println("incorrect");
}
System.out.println(random);
This will take a random double from (0 to .99) and add .5 to make it (.5 to 1.49). It will also cast it to an int, which will make it (0 to 1). The last line is for testing.

for(int i=0;i<100;i++){
System.out.println(((int)(i*Math.random())%2));
}
use mod it will help you!

One more variant
rand.nextInt(2);
As it described in docs it will return random int value between 0 (inclusive) and the specified value (exclusive)

Calculate a value to update JProgressBar

I would like to display the proportion of an initial value in a JProgressBar.
private void updateProgressBars() { //Update the progress bars to the new values.
int p1 = 0, p2 = 1; //Player numbers
double p1Progress = (intPlayer1Tickets/intInitialPlayer1Tickets) * 100;
double p2Progress = (intPlayer2Tickets/intInitialPlayer2Tickets) * 100;
progressbarPlayerTickets[p1].setValue((int) p1Progress);
progressbarPlayerTickets[p1].setString("Tickets left: " + Integer.toString(intPlayer1Tickets));
progressbarPlayerTickets[p2].setValue((int) p2Progress);
progressbarPlayerTickets[p2].setString("Tickets left: " + Integer.toString(intPlayer2Tickets));
}
In this code, the intention was to calculate the percentage of the amount of tickets left a player has. intInitialPlayer1Tickets and intInitialPlayer2Tickets were both set to 50. intPlayer1Tickets and intPlayer2Tickets were then set to their respective initial tickets value (i.e. both set to 50 as well). When I subtract any number from intPlayer1Tickets or intPlayer2Tickets (e.g. intPlayer1Tickets = 49, intInitialPlayer1Tickets = 50), their respective progress bars' value would be set to 0, which is not my intention. Both progress bars have their min and max values set to 0 and 100.
So how would I make it so it would reflect the proportion of tickets left as a percentage?

You are doing integer math and then converting it to a double. In integer math when you divide a number with a number that is bigger, the answer is always 0.
You want to get Java to do your math with floating point numbers rather than with integers. The easiest way to do this is to make your divisor a double.
When you run this code
public class Numbers{
public static void main(String []args){
int five = 5;
int ten = 10;
System.out.println(five/ten);
System.out.println(five/(double)ten);
System.out.println((five/(double)ten)*100);
}
}
You get this as the output
0
0.5
50.0
So, to answer your question what you want is something like this
double p1Progress = (intPlayer1Tickets/(double)intInitialPlayer1Tickets) * 100;
But you'd be just fine using float for this instead of doubles.

Cast the value of intPlayer1Tickets, intPlayer2Tickets, intInitialPlayer1Tickets and intInitialPlayer2Tickets to double before you calculate. As in:
double p1Progress = (((double) intPlayer1Tickets)/((double) intInitialPlayer1Tickets)) * 100;
double p2Progress = (((double) intPlayer2Tickets)/((double) intInitialPlayer2Tickets)) * 100;