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

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

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

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

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

1. Boxing (box IL instruction) and unboxing (unbox IL instruction)
2. Casting through the inhertiance hierarchy (like dynamic_cast<Type> in C++, uses castclass IL 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_XXX method).

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

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

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

The IDataRecord interface also has methods for unboxing the value:

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

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

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