Question

我有为单元测试准备的以下方法，我知道它将始终为每个循环运行，有没有办法摆脱第二个return语句？

public Enums.GYRStatus GetStatusForTransformer(
            string factoryCode, 
            Enums.Technology technology, 
            string transformerType,
            int transformerSize,
            string transformerModel)
{
   fakeStandardsAndSizesFictionary = new Dictionary<Tuple<string,
                                                    Enums.Technology,
                                                    string, int, string>, int>() 
   {
       { Tuple.Create("SELUD", Technology.CVT,"---", 0, ""), 1} };
   }

   foreach (var pair in fakeStandardsAndSizesFictionary)
   {
       if (pair.Key.Item1 == factoryCode &&
          pair.Key.Item2 == technology &&
          pair.Key.Item3 == transformerType &&
          pair.Key.Item4 == transformerSize &&
          pair.Key.Item5 == transformerModel)
           return (Enums.GYRStatus)pair.Value;
   }
   return (Enums.GYRStatus)1; // second return never used
}

Answer 1

您可以替换

return (Enums.GYRStatus)1;

与

throw new InvalidOperationException();

假设永远不会到达这个地方，它在语义上看起来也更正确。

Answer 2

假设(GYRStatus)1是有效的返回值，您可以这样做：

GYRStatus status = GYRStatus.First;
foreach (var pair in fakeStandardsAndSizesFictionary)
{
   if (pair.Key.Item1 == factoryCode &&
   pair.Key.Item2 == technology &&
   pair.Key.Item3 == transformerType &&
   pair.Key.Item4 == transformerSize &&
   pair.Key.Item5 == transformerModel)
   {
      status = (Enums.GYRStatus)pair.Value;
      break;
   }
}
return status;

您还可以返回GYRStatus?以表示未选择任何值：

GYRStatus? status = null;
foreach (var pair in fakeStandardsAndSizesFictionary)
{
   if (pair.Key.Item1 == factoryCode &&
   pair.Key.Item2 == technology &&
   pair.Key.Item3 == transformerType &&
   pair.Key.Item4 == transformerSize &&
   pair.Key.Item5 == transformerModel)
   {
      status = (Enums.GYRStatus)pair.Value;
      break;
   }
}
return status;

或者，您的枚举中也可以包含GYRStatus.None值。这一切都取决于执行的流程。如果有可能最终找不到谓词的值，则返回默认值而不是抛出异常。如果不可能，请抛出。

Answer 3

你可以这样做：

public Enums.GYRStatus GetStatusForTransformer(string factoryCode,   Enums.Technology technology, string transformerType, int transformerSize, string transformerModel)
{
    fakeStandardsAndSizesFictionary = new Dictionary<Tuple<string, Enums.Technology, string, int, string>, int>() 
    {
        {Tuple.Create("SELUD",Technology.CVT,"---",0 ,""),1},
    };

    return fakeStandardsAndSizesFictionary
        .Where(pair =>
            pair.Key.Item1 == factoryCode
                && pair.Key.Item2 == technology
                && pair.Key.Item3 == transformerType
                && pair.Key.Item4 == transformerSize
                && pair.Key.Item5 == transformerModel)
        .Select(pair => (Enums.GYRStatus)pair.Value)
        .First();
}

Answer 4

没有，即使你知道它将永远返回，编译器也不会！

Answer 5

编译器无法知道循环ALLWAYS返回一个值。因此你必须提供那个开关。

Answer 6

public Enums.GYRStatus GetStatusForTransformer(
            string factoryCode, 
            Enums.Technology technology, 
            string transformerType,
            int transformerSize,
            string transformerModel)
{
   Enums.GYRStatus retValue=(Enums.GYRStatus)1;
   fakeStandardsAndSizesFictionary = new Dictionary<Tuple<string,
                                                    Enums.Technology,
                                                    string, int, string>, int>() 
   {
       { Tuple.Create("SELUD", Technology.CVT,"---", 0, ""), 1} };
   }

   foreach (var pair in fakeStandardsAndSizesFictionary)
   {
       if (pair.Key.Item1 == factoryCode &&
          pair.Key.Item2 == technology &&
          pair.Key.Item3 == transformerType &&
          pair.Key.Item4 == transformerSize &&
          pair.Key.Item5 == transformerModel)
           retValue = (Enums.GYRStatus)pair.Value;
   }
   return retValue; 
}

包含foreach语句的方法只有一个返回

6 个答案: