This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Write a truly inclusive random method for javascript
(10 answers)
Closed 9 years ago.
What is the easiest way to include 1 using Math.random function?
Since the default is number between inclusive 1 and exclusive 0.
I am going to use my the rand function a lot of times in my code that is why i need an even probability of getting numbers between 0 and 1 [0,1] , inclusive 0 and inclusive 1.
Thanks!
Edit:
Sorry, big mistake. Exclusive 1, inclusive 0.
Related
This question already has answers here:
How does Java handle integer underflows and overflows and how would you check for it?
(12 answers)
Closed 3 years ago.
I though that the following pieces of code does not compile. However after running it I got unexpected result, I don't understand how it is printing -2 ? can you explain how addition is done here?
int x = 2147483647+2147483647; // it compiles
System.out.print(x); // prints -2
any explain is welcome
Short answer: When Java integers reach their maximum value plus one, they start at their minimum again. It's like going in a circle.
It's like that due to the technical representation of integers as bits. Imagine having 3 bits available to represent a number. You could have the number 111. If you add 1 to it, you'll end up at 1000 but since you only have 3 bits available, it cuts off the first and you end up with 000 which is why you're at the minimum value again after adding 1 to the maximum.
This question already has answers here:
Java Generate Random Number Between Two Given Values [duplicate]
(7 answers)
Closed 5 years ago.
I am using new Random().nextInt(999999) to generate a 6 digit random number. But sometimes it generates 5 digit numbers. And I explicitly have to check if generated number is not 6 digit.
Why this method does that? And is there any other way to be pretty sure it generate only 6 digit random number.
Random().nextInt(999999) generates number between 0 and 999998, so you have ~10% chance to get number smaller than 100000 (5 and less digits).
Try
100000 + Random().nextInt(900000)
you will get number in range 100000 - 999999
Random().nextInt(999999) generates 5 digits number sometimes, because it generates random values between 0 and 999999.
Use the following snippet for generating random numbers in a range:
Random random = new Random();
minimum + random.nextInt(maximum - minimum + 1);
This question already has answers here:
How to work with BIG numbers? [duplicate]
(2 answers)
Large Numbers in Java
(6 answers)
Closed 6 years ago.
How to work with numbers of the order 10^100?Like iterating upto a number of that order,getting sum of squares of their digits;then checking if the sum matches the square of a certain number,if yes sum up those numbers and finally display the sum of those numbers.
You can easily handle this kind of task with strings containing the decimal representation of the numbers.
The sum of squares of the digits of a 100 digit number is bounded by 100*9^2 = 8100: you can use int for that.
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 6 years ago.
I don't know why Java's designers didn't design java.util.Random objects to return random numbers like this:
rand.nextInt(minValue, maxValue);
I don't understand how to pass min and max values to this function. I've seen this recommended:
random.nextInt(max + 1 - (min)) + min
but what does that mean? Does that mean:
random.nextInt(FULL_RANGE + 1) + START_OF_RANGE
For example, if I wanted to generate values between -1 and 1, inclusive, is this how I do it:
int oneOfThreeChoices = rand.nextInt(1 + 1 + 1) - 1;
The API provides sufficient tools for you to complete this task. If you want an random int on the range [a,b], inclusive on both ends, use
int n = random.nextInt(b-a+1) + a
assuming that random is an object previously declared to be of type java.util.Random.
random.nextInt(bound) returns random number from 0 to bound (exclusive). For example random.nextInt(10) returns random number from 0 to 9.
So, if you want to return number from 5 to 15 (inclusive) just say
random.nextInt(11) + 5
Why java designers have not included more convenient API to JDK? I think because the way explained here is pretty simple and obvious.
This question already has answers here:
How do I generate random integers within a specific range in Java?
(72 answers)
Closed 7 years ago.
I'm new to java and was viewing a tutorial and saw nextInt, the guy said that it was supposed to give me a number between -5 and 5 but i have two problems :
(r.nextInt(5 - -5) + -5)
1)In what order does it operate
2)The explanation that i found says that it doesn't accept negatives then why is it supposed to allow me to give me -5?
Thanks a lot.
The answer is unary negative and subtration. Usually, that's written as +. Like
(r.nextInt(5 + 5) + -5)
which is
(r.nextInt(10) - 5)
I assume this is for a Random, in which case it returns [0,10). Which if you subtract 5, is [-5,5).
What your code will do is generate a random integer between 0 and the specified value, and will then subtract 5 from the generated integer.