C# 为什么我不能将 int 拆箱为小数？

Why can't I unbox an int as a decimal?

提问by mezoid

I have an IDataRecord readerthat I'm retrieving a decimal from as follows:

我有一个IDataRecord reader我正在从中检索小数的信息，如下所示：

decimal d = (decimal)reader[0];

For some reason this throws an invalid cast exception saying that the "Specified cast is not valid."

出于某种原因，这会引发无效的强制转换异常，指出“指定的强制转换无效”。

When I do reader[0].GetType()it tells me that it is an Int32. As far as I know, this shouldn't be a problem....

当我这样做时，reader[0].GetType()它告诉我它是一个 Int32。据我所知，这应该不是问题......

I've tested this out by this snippet which works just fine.

我已经通过这个片段测试了这一点，它工作得很好。

int i = 3750;
decimal d = (decimal)i;

This has left me scratching my head wondering why it is failing to unbox the int contained in the reader as a decimal.

这让我摸不着头脑，想知道为什么它无法将读取器中包含的 int 拆箱为十进制。

Does anyone know why this might be occurring? Is there something subtle I'm missing?

有谁知道为什么会发生这种情况？我错过了一些微妙的东西吗？

采纳答案by Mehrdad Afshari

You can only unbox a value type to its original type (and the nullable version of that type).

您只能将值类型拆箱为其原始类型（以及该类型的可为空版本）。

By the way, this is valid (just a shorthand for your two line version):

顺便说一句，这是有效的（只是您的两行版本的简写）：

object i = 4;
decimal d = (decimal)(int)i; // works even w/o decimal as it's a widening conversion

For the reason behind this read this Eric Lippert's blog entry: Representation and Identity

对于这背后的原因，请阅读Eric Lippert 的博客文章：Representation and Identity

Personally, I categorize things done by cast syntax into four different types of operation (they all have different IL instructions):

就我个人而言，我将 cast 语法所做的事情分为四种不同类型的操作（它们都有不同的 IL 指令）：

1. Boxing (boxIL instruction) and unboxing (unboxIL instruction)
2. Casting through the inhertiance hierarchy (like dynamic_cast<Type>in C++, uses castclassIL instruction to verify)
3. Casting between primitive types (like static_cast<Type>in C++, there are plenty of IL instructions for different types of casts between primitive types)
4. Calling user defined conversion operators (at the IL level they are just method calls to the appropriate op_XXXmethod).

1. 装箱（boxIL指令）和拆箱（unboxIL指令）
2. 通过继承层次结构进行转换（如dynamic_cast<Type>在 C++ 中，使用castclassIL 指令进行验证）
3. 原始类型之间的转换（就像static_cast<Type>在 C++ 中，有很多 IL 指令用于原始类型之间的不同类型的转换）
4. 调用用户定义的转换运算符（在 IL 级别，它们只是对适当op_XXX方法的方法调用）。

回答by Guffa

There is no problem in casting an intto decimal, but when you are unboxing an object you have to use the exact type that the object contains.

将intto强制转换没有问题decimal，但是当您拆箱对象时，您必须使用该对象包含的确切类型。

To unbox the intvalue into a decimalvalue, you first unbox it as an int, then cast it to decimal:

要将int值拆箱为decimal值，首先将其拆箱为 int，然后将其转换为十进制：

decimal d = (decimal)(int)reader[0];

The IDataRecord interface also has methods for unboxing the value:

IDataRecord 接口还具有对值进行拆箱的方法：

decimal d = (decimal)reader.GetInt32(0);

回答by user276648

Mehrdad Afshari said it:

Mehrdad Afshari 说：

You can only unbox a value type to its original type (and the nullable version of that type).

您只能将值类型拆箱为其原始类型（以及该类型的可为空版本）。

The thing to realize is that there is a difference between casting and unboxing. jerryjvl had an excellent remark

要意识到的是，cast 和 unboxing 之间存在差异。jerryjvl 发表了精彩的评论

In a sense it's a shame that unboxing and casting syntactically look identical, since they are very different operations.

从某种意义上说，拆箱和强制转换在语法上看起来相同是一种耻辱，因为它们是非常不同的操作。

Casting:

铸件：

int i = 3750; // Declares a normal int
decimal d = (decimal)i; // Casts an int into a decimal > OK

Boxing/Unboxing:

装箱/拆箱：

object i = 3750; // Boxes an int ("3750" is similar to "(int)3750")
decimal d = (decimal)i; // Unboxes the boxed int into a decimal > KO, can only unbox it into a int or int?

回答by Sagar

Here is a simple solution. It takes care of unboxing and then casting to decimal. Worked fine for me.

这是一个简单的解决方案。它负责拆箱，然后转换为十进制。对我来说效果很好。

decimal d = Convert.ToDecimal(reader[0]);  // reader[0] is int