如何在特定范围内生成随机int值？
以下方法存在与整数溢出相关的错误：
randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum = minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.
当前回答
在尝试1中进行以下更改应该可以完成工作-
randomNum = minimum + (int)(Math.random() * (maximum - minimum) );
检查此项以获取工作代码。
其他回答
您可以使用以下方法来执行此操作：
int range = 10;
int min = 5
Random r = new Random();
int = r.nextInt(range) + min;
Use:
Random ran = new Random();
int x = ran.nextInt(6) + 5;
整数x现在是可能结果为5-10的随机数。
我认为这段代码会奏效。请尝试一下：
import java.util.Random;
public final class RandomNumber {
public static final void main(String... aArgs) {
log("Generating 10 random integers in range 1..10.");
int START = 1;
int END = 10;
Random randomGenerator = new Random();
for (int idx=1; idx<=10; ++idx) {
// int randomInt=randomGenerator.nextInt(100);
// log("Generated : " + randomInt);
showRandomInteger(START,END,randomGenerator);
}
log("Done");
}
private static void log(String aMessage) {
System.out.println(aMessage);
}
private static void showRandomInteger(int aStart, int aEnd, Random aRandom) {
if (aStart > aEnd) {
throw new IllegalArgumentException("Start cannot exceed End.");
}
long range = (long)aEnd - (long)aStart + 1;
long fraction = (long) (range * aRandom.nextDouble());
int randomNumber = (int) (fraction + aStart);
log("Generated" + randomNumber);
}
}
如果掷骰子，它将是1到6（而不是0到6）之间的随机数，因此：
face = 1 + randomNumbers.nextInt(6);
另一种选择是使用Apache Commons：
import org.apache.commons.math.random.RandomData;
import org.apache.commons.math.random.RandomDataImpl;
public void method() {
RandomData randomData = new RandomDataImpl();
int number = randomData.nextInt(5, 10);
// ...
}
