c# – Moq设置不适用于方法调用,后跟隐式转换

public interface ITestMe
{
  string TakeInt64(long x);
}

然后运行以下代码：

public void Test()
{
  var mock1 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr1 = x => x.TakeInt64(2);
  Console.WriteLine(expr1);
  mock1.Setup(expr1).Returns("OK");
  var s1 = mock1.Object.TakeInt64(2L); // OK

  var mock2 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr2 = x => x.TakeInt64(DateTime.Today.Year / 1000);
  Console.WriteLine(expr2);
  mock2.Setup(expr2).Returns("OK");
  var s2 = mock2.Object.TakeInt64(2L); // OK

  var mock3 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr3 = x => x.TakeInt64((int)(DayOfWeek)Enum.Parse(typeof(DayOfWeek),"Tuesday"));
  Console.WriteLine(expr3);
  mock3.Setup(expr3).Returns("OK");
  var s3 = mock3.Object.TakeInt64(2L); // OK

  var mock4 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr4 = x => x.TakeInt64(GetInt32());
  Console.WriteLine(expr4);
  mock4.Setup(expr4).Returns("OK");
  //var s4 = mock4.Object.TakeInt64(2L); // MockException,All invocations on the mock must have a corresponding setup.

  var mock5 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr5 = x => x.TakeInt64(int.Parse("2"));
  Console.WriteLine(expr5);
  mock5.Setup(expr5).Returns("OK");
  //var s5 = mock5.Object.TakeInt64(2L); // MockException,All invocations on the mock must have a corresponding setup.

  var mock6 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr6 = x => x.TakeInt64(GetInt32() + 0);
  Console.WriteLine(expr6);
  mock6.Setup(expr6).Returns("OK");
  var s6 = mock6.Object.TakeInt64(2L); // OK

  var mock7 = new Mock<ITestMe>(MockBehavior.Strict);
  Expression<Func<ITestMe,string>> expr7 = x => x.TakeInt64(1 * int.Parse("2"));
  Console.WriteLine(expr7);
  mock7.Setup(expr7).Returns("OK");
  var s7 = mock7.Object.TakeInt64(2L); // OK
}

static int GetInt32()
{
  return 2;
}

在所有七种情况下,我们创建一个表达式树,其中TakeInt64获取int(Int32)而不是long(Int64).但是,众所周知,存在从int到long的隐式转换,它将出现在表达式树中(expr1除外,编译器为我们转换常量).

为什么上面的案例s4和s5不起作用？奇怪的是,正如你所看到的,如果我们在s6和s7的情况下将0或多次加1,那么它是否有效(即使该类型仍为int,隐式转换为long)？

由于案例expr2,请在3000年之前回答.