java 用Java在一定范围内均匀生成安全随机数
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Generate secure random number uniformly over a range in Java
提问by anoopelias
How do I generate a secure uniformrandom number within a range? The range could be between 0 to 100. (The upper bound is not a power of 2).
如何在一个范围内生成一个安全的统一随机数？范围可以在 0 到 100 之间。（上限不是 2 的幂）。
java.security.SecureRandomseems to provide the range 0..2^n.
java.security.SecureRandom似乎提供了 range 0..2^n。
回答by Peter Lawrey
You can do
你可以做
Random rand = new SecureRandom()
// 0 to 100 inclusive.
int number = rand.nextInt(101);
or
或者
// 0 inclusive to 100 exclusive.
int number = rand.nextInt(100);
Note: this is more efficient than say (int) (rand.nexDouble() * 100)as nextDouble() needs to create at least 53-bits of randomness whereas nextInt(100) creates less than 7 bits.
注意：这比(int) (rand.nexDouble() * 100)因为 nextDouble() 需要创建至少 53 位随机性而 nextInt(100) 创建少于 7 位的说法更有效。
回答by Snehal Patel
Try below code snap
尝试下面的代码快照
SecureRandom random = new SecureRandom();
int max=50;
int min =1;
System.out.println(random.nextInt(max-min+1)+min);
