![]() ![]() In your actionPerformed() method, get the selected item and compare it to your random number instance variable: import java.awt. A random bounded integer value can be generated with a single proper. This Java tutorial for beginners explains how to generate random numbers with the random method in the Math class. Add an action listener to your combo box (hopefully you did this already). Generating random floating point values to cast them into integers is inefficient. When you divide with /, the result is the quotient, and the remainder is thrown away. The main difference is, the computer does it in base 2 instead of base 10. ![]() Since there's a 5% chance that the number is below 0.Make the random number an instance variable of your class and populate it in your constructor. Int division in a computer basically is very similar to how you would do long division with paper and pencil to get a quotient and a remainder. So how does this work? Math.random() returns a random double between 0 (including) and 1 (excluding). if a player should have a 5% chance to get an item and a 5% chance to get another, then the player will either get both or none. This can cause (and has for me) strange things, eg. However, if the computer didn't notice any time change (usually with less than a millisecond passing), the two Random objects will return exactly the same thing. If you want to have Numbered from 1 to 100 then its formula will be this-(int)(Math. Any Formula is depended on what you want to accomplish. This is mostly irrelevant because if you have two Random objects, one created today and another one created tomorrow the time will have changed. Answer:Math.random() Returns a double value with a positive sign, greater than or equal to 0.0 and less than 1.0. Whenever you create a new Random object, the random value will depend on the time. Yep, and it's even the better thing to do for the following reason: The ThreadLocalRandom class provides the int nextInt(int origin, int bound) method to get a random integer in a range: // Returns a random int between 1 (inclusive) & 101 (exclusive) int randomInt ThreadLocalRandom.current(). Since there's a 5% chance that the number is below 0.05, you can use that for your code: ![]() This will guarantee that even when calling it a hundred times in the same nanosecond, it will always return another number. If you use Math.random() however, there'll be a single Random instance for all of your calculations. if a player should have a 5% chance to get an item and a 5% chance to get another, then the player might either get both or none. Still, doing this is considered bad practice and other implementations might not have one). However, if the computer didn't notice any time change (usually when less than a millisecond passed), the two Random objects might return exactly the same thing (edit 2019: Not 100% accurate, there's a seed uniquifier in the Java default Random implementation to prevent this. This is because of the cast to integer (int) Math.random() You see Math.random() gives a random number between 0.0 and 1.0 when you cast to an integer you are effectively saying - ' Throw out all the decimal places' So if you had 0.5 doing (int) Math.random() would give you 0. This is usually irrelevant because if you have two Random objects, one created today and another one created tomorrow the time will have changed. Whenever you create a new Random object, the returned value will depend on the current time. Click to expand.Yep, and it's even the better thing to do for the following reason: ![]()
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