This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 3 years ago.
I would like to get a random value between 1 to 50 in Java.
How may I do that with the help of Math.random();?
How do I bound the values that Math.random() returns?
The first solution is to use the java.util.Random class:
import java.util.Random;
Random rand = new Random();
// Obtain a number between [0 - 49].
int n = rand.nextInt(50);
// Add 1 to the result to get a number from the required range
// (i.e., [1 - 50]).
n += 1;
Another solution is using Math.random():
double random = Math.random() * 49 + 1;
or
int random = (int)(Math.random() * 50 + 1);
int max = 50;
int min = 1;
1. Using Math.random()
double random = Math.random() * 49 + 1;
or
int random = (int )(Math.random() * 50 + 1);
This will give you value from 1 to 50 in case of int
or 1.0 (inclusive) to 50.0 (exclusive) in case of double
Why?
random() method returns a random
number between 0.0 and 0.9..., you
multiply it by 50, so upper limit
becomes 0.0 to 49.999... when you add 1, it becomes 1.0 to 50.999..., now when you truncate to int, you get 1 to 50. (thanks to #rup in comments). leepoint's awesome write-up on both the approaches.
2. Using Random class in Java.
Random rand = new Random();
int value = rand.nextInt(50);
This will give value from 0 to 49.
For 1 to 50: rand.nextInt((max - min) + 1) + min;
Source of some Java Random awesomeness.
Related
This question already has answers here:
How to get a random between 1 - 100 from randDouble in Java?
(6 answers)
Closed 2 years ago.
So I am trying to make 100 random integers from 1-100 print out. I can make doubles of this but not integers. Please show me how to make this into integers, here is the code.
for(i = 1; i <= 100; i++) {
double r = Math.random();
double in = r * 100;
System.out.print(in + ", ");
}
Please show me what I messed up or if I need to use an alternate technique for this
You can use Random object instead. from java.util package
Random random = new Random();
int value = random.nextInt(100);
Use ThreadLocalRandom:
for (int i = 1; i <= 100; i++) {
System.out.print(ThreadLocalRandom.current().nextInt(100) + ", ");
}
Import the class java.util.Random
Make the instance of the class Random, i.e., Random rand = new Random()
Invoke one of the following methods of rand object:
nextInt(upperbound) generates random numbers in the range 0 to upperbound-1.
nextFloat() generates a float between 0.0 and 1.0.
nextDouble() generates a double between 0.0 and 1.0
For example if you want to generate a random number between 1 to 100 the the code will be.
import java.util.Random;
Random r = new Random();
Int number = r.nextInt(100);
So variable number will be a random number between 1 to 100
Whats wrong is the statement double in = r * 100;
It must be
int b = (int) (Math.random()*100)%100; //range [0-99]
Just add 1 to b if you strictly want b in the range [1-100]
The following code is only producing a 0 ;-;
What am I doing wrong?
public class RockPaperSci {
public static void main(String[] args) {
//Rock 1
//Paper 2
//Scissors 3
int croll =1+(int)Math.random()*3-1;
System.out.println(croll);
}
}
Edit, Another Poster suggested something that fixed it.
int croll = 1 + (int) (Math.random() * 4 - 1);
Thanks, everyone!
You are using Math.random() which states
Returns a double value with a positive sign, greater than or
equal to 0.0 and less than 1.0.
You are casting the result to an int, which returns the integer part of the value, thus 0.
Then 1 + 0 - 1 = 0.
Consider using java.util.Random
Random rand = new Random();
System.out.println(rand.nextInt(3) + 1);
Math.random() generates double values between range - [0.0, 1.0). And then you have typecasted the result to an int:
(int)Math.random() // this will always be `0`
And then multiply by 3 is 0. So, your expression is really:
1 + 0 - 1
I guess you want to put parenthesis like this:
1 + (int)(Math.random() * 3)
Having said that, you should really use Random#nextInt(int) method if you want to generate integer values in some range. It is more efficient than using Math#random().
You can use it like this:
Random rand = new Random();
int croll = 1 + rand.nextInt(3);
See also:
Math.random() versus Random.nextInt(int)
All our mates explained you reasons of unexpected output you got.
Assuming you want generate a random croll
Consider Random for resolution
Random rand= new Random();
double croll = 1 + rand.nextInt() * 3 - 1;
System.out.println(croll);
public static double random()
Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0. Returned values are chosen pseudorandomly with (approximately) uniform distribution from that range.
int croll =1+(int)Math.random()*3-1;
eg
int croll =1+0*-1;
System.out.println(croll); // will print always 0
This question already has answers here:
Generate a random double in a range
(7 answers)
Closed 9 years ago.
I'm a Java noob trying to generate a random double between -10 and 10 inclusive. I know with ints I would do the following:
Random r = new Random();
int i = -10 + r.nextInt(21);
However, with doubles, this doesn't work:
Random r = new Random();
double i = -10 + r.nextDouble(21);
Can someone please explain what to do in the case of doubles?
Try this:
Random r = new Random();
double d = -10.0 + r.nextDouble() * 20.0;
Note: it should be 20.0 (not 21.0)
Try to use this for generate values in given range:
Random random = new Random();
double value = min + (max - min) * random.nextDouble();
Or try to use this:
public double doubleRandomInclusive(double max, double min) {
double r = Math.random();
if (r < 0.5) {
return ((1 - Math.random()) * (max - min) + min);
}
return (Math.random() * (max - min) + min);
}
Returns the next pseudorandom, uniformly distributed double value between 0.0 and 1.0 from this random number generator's sequence.
nextDouble doesn't take a parameter, you just need to multiply it by whatever your range is. Or more generally:
minimum + (maximum - minimum) * r.nextDouble();
There is no nextDouble(int) method in Random. Use this:
double i = -10.0 + r.nextDouble() * 20.0;
API reference
I need to make a random number between 1 and 20, and based on that number (using "If - Then" statements), I need to set the image of an ImageView.
I know that in Objective-C, it goes like this:
int aNumber = arc4Random() % 20;
if (aNumber == 1) {
[theImageView setImage:theImage];
}
How can I do this in Java? I have seen it done this way, but I do not see how I can set the range of numbers (1-20, 2-7, ect).
int aNumber = (int) Math.random()
Docs are your friends
Random rand = new Random();
int n = rand.nextInt(20); // Gives n such that 0 <= n < 20
Documentation:
Returns a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive), drawn from this random number generator's sequence.
Thus, from this example, we'll have a number between 0 and 19
Math.random() returns an double from [0,1[.
Random.nextInt(int) returns an int from [0, int[.
You can try:
int aNumber = (int) (20 * Math.random()) + 1;
or
Random rand = new Random();
int n = rand.nextInt(20) + 1;
You can use Math.random() to generate a double between 0 and 1 non-inclusive. Android Javadoc here.
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 4 years ago.
I am trying to generate random integers over the range (-32768, 32767) of the primitive data type short. The java Random object only generates positive numbers. How would I go about randomly creating numbers on that interval? Thanks.
You random on (0, 32767+32768) then subtract by 32768
Random random=new Random();
int randomNumber=(random.nextInt(65536)-32768);
public static int generatRandomPositiveNegitiveValue(int max , int min) {
//Random rand = new Random();
int ii = -min + (int) (Math.random() * ((max - (-min)) + 1));
return ii;
}
Generate numbers between 0 and 65535 then just subtract 32768
This is an old question I know but um....
n=n-(n*2)
([my double-compatible primitive type here])(Math.random() * [my max value here] * (Math.random() > 0.5 ? 1 : -1))
example:
// need a random number between -500 and +500
long myRandomLong = (long)(Math.random() * 500 * (Math.random() > 0.5 ? 1 : -1));
(Math.floor((Math.random() * 2)) > 0 ? 1 : -1) * Math.floor((Math.random() * 32767))
In case folks are interested in the double version (note this breaks down if passed MAX_VALUE or MIN_VALUE):
private static final Random generator = new Random();
public static double random(double min, double max) {
return min + (generator.nextDouble() * (max - min));
}