为什么c#中的Random构造函数在方法的最后指定Seed参数？

Question

我在这里检查了.Net的Random类的源代码.令我惊讶的是最后一行

  public Random(int Seed) {
    int ii;
    int mj, mk;

    //Initialize our Seed array.
    //This algorithm comes from Numerical Recipes in C (2nd Ed.)
    int subtraction = (Seed == Int32.MinValue) ? Int32.MaxValue : Math.Abs(Seed);
    mj = MSEED - subtraction;
    SeedArray[55]=mj;
    mk=1;
    for (int i=1; i<55; i++) {  //Apparently the range [1..55] is special (Knuth) and so we're wasting the 0'th position.
      ii = (21*i)%55;
      SeedArray[ii]=mk;
      mk = mj - mk;
      if (mk<0) mk+=MBIG;
      mj=SeedArray[ii];
    }
    for (int k=1; k<5; k++) {
      for (int i=1; i<56; i++) {
    SeedArray[i] -= SeedArray[1+(i+30)%55];
    if (SeedArray[i]<0) SeedArray[i]+=MBIG;
      }
    }
    inext=0;
    inextp = 21;
    Seed = 1;
  }

在方法结束时分配参数的目的是什么？

Answer 1

由于Seed参数未被传递ref并且int是值类型,因此最后一行无论如何都没有效果.

感谢@Alex K指出这一点,在检查了他们在C中的Numerical Recipes中调整的算法后,他们确实复制并粘贴了最后一行:

if ( * idum < 0 || iff == 0) {
  Initialization.
  iff = 1;
  mj = labs(MSEED - labs( * idum));
  Initialize ma[55] using the seed idum and the
  mj %= MBIG;
  large number MSEED.
  ma[55] = mj;
  mk = 1;
  for (i = 1; i <= 54; i++) {
    ii = (21 * i) % 55;
    ma[ii] = mk;
    mk = mj - mk;
    if (mk < MZ) mk += MBIG;
    mj = ma[ii];
  }
  for (k = 1; k <= 4; k++)
    (i = 1; i <= 55; i++) {
    ator.”
    ma[i] -= ma[1 + (i + 30) % 55];
    if (ma[i] < MZ) ma[i] += MBIG;
  }
  inext = 0;
  inextp = 31;
  * idum = 1;
}