CODEWITHSUNDEEP
java
Michel Tamer
Michel Tamer

Reputation: 341

Differences in random functions

I am creating a random number from 1-100, I was looking at some Stackoverflow questions to look for the proper way and I got confused by the many different suggestions. What is the difference between using this:

    int random= (int)(Math.random()*((100-1)+1));

this:

int random= (int)(Math.random()*(100);

and this:

    int random= 1+ (int)(Math.random()*((100-1)+1));

Upvotes: 3

Views: 216

Answers (3)

matt
matt

Reputation: 2449

int random = (int)(Math.random()*(x);

This sets random equal to any integer between 0 and x - 1. 

int random = 1 + (int)(Math.random()*(x);

Adding 1 to the overall expression simply changes it to any integer between 1 and x.

(int)(Math.random()*((100-1)+1))

is redundant and equivalent to 

(int)(Math.random()*(100)

So take note that:

1 + (int)(Math.random()*(x) returns an int anywhere from 1 to x + 1

but

(int)(Math.random()*(x + 1) returns an int anywhere from 0 to x + 1.

Upvotes: 5

Christian Tapia
Christian Tapia

Reputation: 34166

The first is equivalent to the second. Both will give a random integer between 0 and 99 (inclusive, because Math.random() returns a double in the range [0, 1)). Note that (100-1)+1 is equivalent to 100.

The third, will give an integer between 1 and 100 because you are adding 1 to the result above, i.e. 1 plus a value in the range [0, 100), which results in the range [1, 101).

Upvotes: 1

Elliott Frisch
Elliott Frisch

Reputation: 201497

I recommend that you use Random and nextInt(100) like so,

java.util.Random random = new java.util.Random();
// 1 to 100, the 100 is excluded so this is the correct range.
int i = 1 + random.nextInt(100);

it has the added benefit of being able to swap in a more secure random generator (e.g. SecureRandom). Also, note that you can save your "random" reference to avoid expensive (and possibly insecure) re-initialization.

Upvotes: 3

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