c# - 将数学公式应用到新列表时如何保持精度？

Question

这个列表有 5 个对象，如果我用第二个对象减去第一个对象:[0]-[1] 并将其放入一个新列表中，它就可以工作。但看看这个:[2]-[3], [4]-[5] =？请注意，我有 [4] 作为我的第 5 项，但我没有“第 6 项”:[5] 因为这个列表只有 5 项，我要减去什么？我会收到错误/异常吗？

        var wqat = 1.1;
        var rat = .2;
        var eat = .8;
        var baat = 1.2;
        var baat2 = 1.8;

        List<double> test = new List<double>(); 
        List<double> theOneList = new List<double>();

        theOneList.Add(wqat);
        theOneList.Add(rat);
        theOneList.Add(eat);
        theOneList.Add(baat);
        theOneList.Add(baat2);

        theOneList = theOneList.OrderByDescending(z => z).ToList();

        for (int i = 0; i < 5; i++)
        {
            test.Add(theOneList[i] - theOneList[i + 1]);
            Console.WriteLine(test[i]);
        }

我在尝试减法时遇到系统超出范围异常，我尝试实现的一个可能的解决方案是，如果列表是“奇数”，则只需将“0”添加到列表中，它就会使其成为偶数对象，这样我就可以一次平静地减去 2 个数字。

Answer 1

上述评论摘要:

  var wqat = 1.1;
  var rat = .2;
  var eat = .8;
  var baat = 1.2;
  var baat2 = 1.8;

  // Add's are hard to read
  List<double> theOneList = new List<double>() {
    wqat, rat, eat, baat, baat2 
  };

  // Inplace sorting, "-" for the descending order
  theOneList.Sort((x, y) => -x.CompareTo(y));

  // Or (worse) Linq, do not forget to assign the result
  // theOneList = theOneList.OrderByDescending(z => z).ToList();

  List<double> test = new List<double>(); 

  // No magic numbers (i.e. 4) - no pesky out of range exceptions
  for (int i = 0; i < theOneList.Count - 1; ++i)
    test.Add(theOneList[i] - theOneList[i + 1]);

  // Output (mix output and algorthim is not a good idea)
  Console.Write(String.Join(Environment.NewLine, test));