Java的SecureRandom循环

问题描述

PRNG通常有一个循环，之后生成的随机数会重复。如下创建的SecureRandom的实例时什么的Java的SecureRandom的周期：

SecureRandom的随机= SecureRandom.getInstance（ SHA1PRNG）;

解决方案

我有点困惑。我查看了openjdk的sun.security.provider.SecureRandom的代码。这里内部状态更新如下：

  digest.update（state）; 
 output = digest.digest（）; 
 updateState（state，output）; 
  

[...]

  private static void updateState（byte [] state，byte [] output）{
 int last = 1; 
 int v = 0; 
 byte t = 0; 
 boolean zf = false; 
 
 // state（n + 1）=（state（n）+ output（n）+ 1）％2 ^ 160; 
 for（int i = 0; i< state.length; i ++）{
 //添加两个字节
v =（int）state [i] +（int）output [i] +最后; 
 //结果是低8位
 t =（字节）v; 
 //存储结果。检查状态冲突。 
 zf = zf | （state [i]！= t）; 
 state [i] = t; 
 //高8位进位。存储下一次迭代。 
 last = v>> 8; 
} 
 
 //确保至少有一位变化！ 
 if（！zf）
 state [0] ++; 
} 
  

没有计数器递增，但内部状态只是用输出更新。 / p>

PRNGs usually have a cycle after which the generated random numbers do repeat. What's the cycle of SecureRandom of Java when the instance of SecureRandom is created as follows:

SecureRandom random = SecureRandom.getInstance("SHA1PRNG");

解决方案

I'm a bit confused. I had a look into the code of sun.security.provider.SecureRandom of the openjdk. Here the internal state is updated as follows:

digest.update(state);
output = digest.digest();
updateState(state, output);

[...]

private static void updateState(byte[] state, byte[] output) {
    int last = 1;
    int v = 0;
    byte t = 0;
    boolean zf = false;

    // state(n + 1) = (state(n) + output(n) + 1) % 2^160;
    for (int i = 0; i < state.length; i++) {
        // Add two bytes
        v = (int)state[i] + (int)output[i] + last;
        // Result is lower 8 bits
        t = (byte)v;
        // Store result. Check for state collision.
        zf = zf | (state[i] != t);
        state[i] = t;
        // High 8 bits are carry. Store for next iteration.
        last = v >> 8;
    }

    // Make sure at least one bit changes!
    if (!zf)
       state[0]++;
}

No counter is incremented but the internal state is simply updated with the output.

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