ValidPairs.remove(new Pair(duckDudeX, duckDudeY)) My version uses a list of valid square coordinates, so it requires a bit of initialization: List validPairs = new ArrayList(49)
This is also the basic idea in a couple of the other answers. The other option is you exclude the forbidden square already when picking the square. What is not so satisfactory about it is that even though it will eventually terminate, we know no upper bound on the number of iterations it will take. And again until you get a different position: int x If you hit the duckDude position, you pick again.
The simple way is, you pick the two random integers.
Greenfoot random number code#
Methods, how can I make it generate a random location that excludes 5 and 5 together? The following code will move an object forward four spaces and then generate a random number between zero and 99 inclusive. So, is there a way I can generate these random ints, using the public int getRandomNumber(int start,int end) Whenever the Wombat hits a wall, reset the number of steps to zero. Note: The wombat will still have to turn left if it hits a wall before taking 5 steps. Task 2) Modify the Wombat class so that every 4 steps that it makes it will turn left. As you have probably seen, this poses quite a problem: if theDuckDude and a coin actor are generated on the same square, the game won't work properly: Task 1) Modify the Wombat class so that it takes 30 steps and stops. This could be very dificult to find if your programm doesn't work bekause of that. But Moritz, you should be carefull with this: Greenfoot.getRandomNumber (50) is a random number from 0 to 49. Obviously, this code shows that when my theDuckDude character touches the coin actor, the coin is removed. I always just write Greenfoot.getRandomNumber (number)+ohterNumber but this way is better.
Here is a snippet of the code from my theDuckDude character: Actor actor = getOneIntersectingObject(coin.class) So, as you can see, the three coins and duck character will appear on the screen in random locations. MainCharacter theDuckDude = new MainCharacter () ĪddObject(coin1, getRandomNumber(1, 7), getRandomNumber(1, 7)) ĪddObject(coin2, getRandomNumber(1, 7), getRandomNumber(1, 7)) ĪddObject(coin3, getRandomNumber(1, 7), getRandomNumber(1, 7)) Create a new world with 600x400 cells with a cell size of 1x1 pixels. * Constructor for objects of class FirstLevel. Public int getRandomNumber(int start,int end) I have written this code: public class FirstLevel extends World I've been working on a project in Greenfoot, with Java, where the main character is on a 7x7 screen with 3 coins. Int normal = Greenfoot.getRandomNumber(end-start+1) I'm not getting errors as it is compiling fine, but when i run it i have issues.EDIT: This is a different question to How do I generate random integers within a specific range in Java? because I requested the solution uses the methods public int getRandomNumber(int start,int end) Invincible invincible = new Invincible() ĪddObject(invincible, Greenfoot.getRandomNumber(getWidth()), Greenfoot.getRandomNumber(getHeight())) ĪddObject(storm, Greenfoot.getRandomNumber(getWidth()), Greenfoot.getRandomNumber(getHeight())) * Constructor for objects of class Level_One.ĪddObject(ninad, getWidth()/2, getHeight()/2) ĪddObject(fail, Greenfoot.getRandomNumber(getWidth()), Greenfoot.getRandomNumber(getHeight())) ĪddObject(new Fail(), Greenfoot.getRandomNumber(getWidth()), Greenfoot.getRandomNumber(getHeight())) ĪddObject(new APlus(), Greenfoot.getRandomNumber(getWidth()), Greenfoot.getRandomNumber(getHeight())) * Write a description of class Level_One here. I'm somewhat new to Java so it's probably something simple. For some reason, the object will be created every time the counter reaches 600. So I'm generating a random number every time a counter reaches 600, and then testing if that randomly generated number is equal to 1, and if it is, it creates an object. I'm making a game in Greenfoot and I'm having an issue. I know this might already have been answered, but all the places where i found it, it wouldn't work properly.
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