What is the inner difference between (MyType)SomeObj.Property1 and (MyType)(SomeObj.Property1) in C# ?

Question

It's probably very lame question, but I found no references in C# specification about round brackets. Please point me to spec or msdn if answer on that question will be obvious.

What is the inner difference between (MyType)SomeObj.Property1 and (MyType)(SomeObj.Property1) in C# ?

AFAIK, in first case ( (x)SomeObj.Property1 cast ) - it will be the reference of concrete type (MyType) to Property1. In second case, such reference will execute get accessor SomeObj.get_Property1. And it eventually could lead to subtle errors if get accessor have any side effects (and its often - do have ones)

Could anyone point me to exact documentation where such behaviour specified ?

Updated: Thank you for pointing. And I deeply apologize about such dumb question - after posting this question, I found a typo in example I fiddle with and thus realized that second case behaviour was not based on code I tried to compile, but on previously compiled completely different code. So my question was initially based on my own blindness ...

Answer 1

There is absolutely no difference. The . operator binds more tightly than the typecast operator, so the extra parentheses make no difference. See here for details of operator precedence; the operators in question are in the first two groups.

Answer 2

They are equivalent. This is determined by the operator precedence rules in the C# language, chapter 7.2.1 in the C# Language Specification:

The . operator is at the top in this list, the cast operator is the 2nd in the list. The . operator "wins". You will have use parentheses if you need the cast because Property1 is a property of the MyType class:

 ((MyType)SomeObj).Property1