java 在java中生成短随机数？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/10189016/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
generate short random number in java?
提问by waqas
I want to generate a random number of type short exactly like there is a function for integer type called Random.nextInt(134116). How can I achieve it?
我想生成一个 short 类型的随机数，就像有一个名为 Random.nextInt(134116) 的整数类型函数一样。我怎样才能实现它？
回答by luketorjussen
There is no Random.nextShort()method, so you could use
没有Random.nextShort()方法，所以你可以使用
short s = (short) Random.nextInt(Short.MAX_VALUE + 1);
The +1 is because the method returns a number up to the number specified (exclusive). See here
+1 是因为该方法返回一个数字，直到指定的数字（不包括）。看这里
This will generate numbers from 0 to Short.MAX_VALUE inclusive (negative numbers were not requested by the OP)
这将生成从 0 到 Short.MAX_VALUE 的数字（OP 未请求负数）
回答by Peter Lawrey
The most efficient solution which can produce all possible short values is to do either.
可以产生所有可能的短值的最有效的解决方案是执行任一操作。
short s = (short) random.nextInt(1 << 16); // any short
short s = (short) random.nextInt(1 << 15); // any non-negative short
or even faster
甚至更快
class MyRandom extends Random {
public short nextShort() {
return (short) next(16); // give me just 16 bits.
}
public short nextNonNegativeShort() {
return (short) next(15); // give me just 15 bits.
}
}
short s = myRandom.nextShort();
回答by Skippy Fastol
Java shorts are included in the -32 768 → +32 767 interval.
Java Shorts 包含在 -32 768 → +32 767 区间内。
why wouldn't you perform a
你为什么不表演一个
Random.nextInt(65536) - 32768
and cast the result into a shortvariable ?
并将结果转换为一个短变量？
回答by assylias
How about short s = (short) Random.nextInt();? Note that the resulting distribution might have a bias. The Java Language Specification guarantees that this will not result in an Exception, the int will be truncated to fit in a short.
怎么样short s = (short) Random.nextInt();？请注意，结果分布可能存在偏差。Java 语言规范保证这不会导致异常，int 将被截断以适应 short。
EDIT
编辑
Actually doing a quick test, the resulting distribution seems to be uniformly distributed too.
实际上做一个快速测试，结果分布似乎也是均匀分布的。
回答by Tudor
Simply generate an int like:
只需生成一个 int 像：
short s = (short)Random.nextInt(Short.MAX_VALUE);
The generated intwill be in the value space of short, so it can be cast without data loss.
生成的int将在 的值空间中short，因此可以在不丢失数据的情况下进行转换。
