This question already has answers here:
How to test if a double is an integer
(18 answers)
Closed 8 years ago.
There are several ways to decide if a value is an integer or not.
All of the ways i know are using the divide operation.
Which method is the fastest and:
Is there a method to do this without doing floating point operations?
EDIT: For clarification, my program is only dealing with integer and double values. Here is some of my code:
for (int i = 6;; i++) {
for (int j = i - 1; j > 0; j--) {
double number1 = i;
double number2 = j;
double d = number1 / number2;
int help = (int)d;
if( (d - help) == 0.0){
System.out.println("Whole number found");
}
}
}
Note that int / int will also be an int due to the rules of integral division. You need to promote one of them to floating point or use the idiom 1.0 * a / b if you want to retain a remainder.
If you have a floating point number, f, then
java.lang.Math.floor(f) - f == 0 is probably the best way.
It avoids an intermediate cast to int which can overflow.
This will do what you want!
String str = "1";
try {
Integer.parseInt(str);
} catch (NumberFormatException e) {
System.err.println("Not a number");
}
Related
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
How to divide 1 by 25(1/25) and get result .04 in java [duplicate]
(4 answers)
Closed 1 year ago.
I am trying to write a function to check a T score value and populate half of a 5x5 array.
public void calcTScores()
{
double temp = 0;
double tempSp = 0;
int n = 24;
this.tScores = new String [5][5];
for (int i = 0; i< this.Headers.length; i++){
for (int j = 0; j<this.Headers.length; j++)
{
if(i < j)
{
tempSp += (n-1)*this.SD[i] * this.SD[i] + (n-1)*this.SD[j] * this.SD[j];
tempSp = tempSp/(n+n-2);
tempSp = Math.sqrt(tempSp);
temp = tempSp * Math.sqrt(0.0833);
System.out.println(Math.sqrt(1/12));
temp = ((this.Mean[i] - this.Mean[j])/temp);
if(temp > 2.25 || temp< -2.25)
{
this.tScores[i][j] = "Y";
}
else
{
this.tScores[i][j] = "N";
}
temp = 0;
tempSp = 0;
}
}
}
}
Any idea why Math.sqrt(0.0833) and Math.sqrt(1/12) would evaluate to different values?
The T score when I add the 1/24 and 1/24 value and take the sqrt keeps evaluating to zero but when I plug in the actual decimal it gives me the answer I would expect
Any ideas why this is occuring?
1/12==0 as per integer division
There's nothing wrong with Math.sqrt. You're passing it the number zero.
Math.sqrt(1/12)
1/12 is a division operation on two integers (namely, 1 and 12), so it's going to produce an integer result. 12 goes into 1 zero times, so the integer result is zero. Thus, your expression is
Math.sqrt(0)
Consider
Math.sqrt(1.0/12.0)
Math.sqrt(1/12d)
Cast 1/12 to a double by casting one operand to a double.
double formula1, formula2;
int plus;
int VALUE = 10000;
private void processFormula2()
{
for (int k = 0; k <= VALUE; k++) {
if (k % 2 != 0) {
if (plus % 2 == 0) {
double math = 1/k;
formula2 += math;
System.out.println("Getting Formula: "+ formula2);
plus++;
} else {
formula2 -= 1/k;
plus++;
}
// System.out.println("Term: " + formula2);
}
}
}
I am trying to get my formula to print out the result of Pi based off this formula that my teacher gave us. But for some reason it just returns 1.0, not really sure why. Any help or suggestions would be appreciated :)
Here's the problem:
double math = 1/k;
and
formula2 -= 1/k;
k is an int variable, so the JVM won't never return a decimal number in this statement. It will take only two possible values: 0 (if k > 1) or 1 (if k == 1) because the JVM performs the division before promoting the result to double.
Try this:
formula2 -= 1/(double)k;
Take a look at Numeric Promotions
Firstly, there are multiple errors with variable declaration.
double math = 1/k;will not truly work in Java due to how integer division is handled. You must either cast '1' to a double like double math = (double)1/k; or specify that you are using mixed mode arithmetic by using double math = 1.0/k;. This is also a problem for your formula2 variable (Along with you should always initialize your variables like formula1, formula2, and plus). You must also do the same thing with formula2 -= 1/k;.
Secondly, we have no idea what you are setting those variables to in the first place, nor do we have any test cases to compare to.
This question already has answers here:
The literal xyz of type int is out of range
(5 answers)
Closed 8 years ago.
The code is supposed to give back the biggest prime number.
More about the task here: https://projecteuler.net/problem=3
int checkFactors(double na) {
long n = (long) na;
int biggestPrimeFactor = 0;
for (int i = 1; i < n; i++)
if (n % i == 0 && isPrimFaktor(i) && i > biggestPrimeFactor)
biggestPrimeFactor = i;
return biggestPrimeFactor;
}
boolean isPrimeFactor(int n) {
int length= 0;
for (int i = n; i > 0; i--)
if (n % i == 0)
length++;
if (length== 2)
return true;
return false;
}
I decided to make the parameter of checkFactors() a double because I tried to test why my code didn't work properly.
System.out.println(checkFactors(13195));
works and returns "29".
However, System.out.println(checkFactors(600851475143));
does not work,
"600851475143 of type int is out of range".
System.out.println(checkFactors(600851475143.0));
does compile but gives me after a couple of seconds an ArithmeticException.
600851475143 of type int is out of range
This number is bigger than int can store. Appending .0 to the number converts the number into a double which can represent that number
Instead of .0 you can do checkFactors(600851475143d) which ensure the number is a double and not an int
Use long as a data type for na and also biggestPrimeFactor. The values are too large for storing in an int variable.
Try to make Your parameter back to long and make letter L after your large number like this 600851475143L, I think it will work
I am having trouble with floating points. A the double . 56 in Java, for example, might actually be stored as .56000...1.
I am trying to convert a decimal to a fraction. I tried to do this using continued fractions
Continuous Fractions
but my answers using that method were inaccurate due to how to computer stored and rounded decimals.
I tried an alternative method:
public static Rational rationalize(double a){
if(a>= 1){
//throw some exception
}
String copOut = Double.toString(a);
int counter = 0;
System.out.println(a);
while(a%1 != 0 && counter < copOut.length() - 2){
a *= 10;
counter++;
}
long deno = (long)Math.pow(10,counter);//sets the denominator
Rational frac = new Rational((long)a,deno)//the unsimplified rational number
long gcd = frac.gcd();
long fnum = frac.getNumer();//gets the numerator
long fden = frac.getDenom();//gets the denominator
frac = new Rational(fnum/gcd, fden/gcd);
return frac;
}
I am using the string to find the length of the decimal to determine how many time I should multiply by 10. I later truncate the decimal. This gets me the right answer, but it does not feel like the right approach?
Can someone suggest the 'correct' way to do this?
Actually you are doing great.. But this will fail if the Input is something about 11.56. Here you need to to do copOut.length() - 3.
To make it dynamic use String#split()
String decLength = copOut.split("\\.")[1]; //this will result "56" (Actual string after decimal)
Now you just need to do only
while(a%1 != 0 && counter < decLength.length()){
a *= 10;
counter++;
}
If you want to remove the loop then use
long d = (long)Math.pow(10,decLength.length());
a=a*d;
This question already has answers here:
Return first digit of an integer
(25 answers)
Closed 5 years ago.
I am just learning Java and am trying to get my program to retrieve the first digit of a number - for example 543 should return 5, etc. I thought to convert to a string, but I am not sure how I can convert it back? Thanks for any help.
int number = 534;
String numberString = Integer.toString(number);
char firstLetterChar = numberString.charAt(0);
int firstDigit = ????
Almost certainly more efficient than using Strings:
int firstDigit(int x) {
while (x > 9) {
x /= 10;
}
return x;
}
(Works only for nonnegative integers.)
int number = 534;
int firstDigit = Integer.parseInt(Integer.toString(number).substring(0, 1));
firstDigit = number/((int)(pow(10,(int)log(number))));
This should get your first digit using math instead of strings.
In your example log(543) = 2.73 which casted to an int is 2.
pow(10, 2) = 100
543/100 = 5.43 but since it's an int it gets truncated to 5
int firstDigit = Integer.parseInt(Character.toString(firstLetterChar));
int number = 534;
String numberString = "" + number;
char firstLetterchar = numberString.charAt(0);
int firstDigit = Integer.parseInt("" + firstLetterChar);
Integer.parseInt will take a string and return a int.
This example works for any double, not just positive integers and takes into account negative numbers or those less than one. For example, 0.000053 would return 5.
private static int getMostSignificantDigit(double value) {
value = Math.abs(value);
if (value == 0) return 0;
while (value < 1) value *= 10;
char firstChar = String.valueOf(value).charAt(0);
return Integer.parseInt(firstChar + "");
}
To get the first digit, this sticks with String manipulation as it is far easier to read.
int number = 534;
int firstDigit = number/100;
( / ) operator in java divide the numbers without considering the reminder so when we divide 534 by 100 , it gives us (5) .
but if you want to get the last number , you can use (%) operator
int lastDigit = number%10;
which gives us the reminder of the division , so 534%10 , will yield the number 4 .
This way might makes more sense if you don't want to use str methods
int first = 1;
for (int i = 10; i < number; i *= 10) {
first = number / i;
}