为什么在强制转换运算符中将 double 隐式转换为 int？

问：

这是我以前没有注意到的奇怪行为。我预计无法从 a 强制转换，但该值被隐式转换为 an，我默默地丢失了小数部分。我真的不能在不从 s 转换的情况下从 s 转换吗？doubledoubleintintdouble

readonly struct TestDecimal {
   public decimal Value { get; }

   TestDecimal(decimal value) => 
      Value = value;

   public static explicit operator TestDecimal(int x) => new(x);
}

...

// these assertions fail, but really this shouldn't even compile
[TestMethod]
public void Test_CastFromDouble() {
   var value = (TestDecimal)12.34; // shouldn't compile
   Assert.AreEqual(12.34m, value.Value); // assertion failed, actual value: 12.0m
}

[TestMethod]
public void Test_CastFromDecimal() {
   var value = (TestDecimal)12.34m; // shouldn't compile
   Assert.AreEqual(12.34m, value.Value); // assertion failed, actual value: 12.0m
}