Strange IL code from simple C# expression

[Anonymous]

The value of i will be 5 because the i++ increments the value AFTER the assignment. If you want i to be 6 then you can do it like this: int i = 5; i = ++i; The method above increments the value BEFORE the assignment. By the way, this is not just a C# thing, any C syntax language shoould behave this way (incl. java, javascript, C++ etc..) Si

[Andrew Kirillov]

Anonymous wrote: By the way, this is not just a C# thing, any C syntax language shoould behave this way (incl. java, javascript, C++ etc..) Yes, maybe other languages also should, but C++ produces 6. (managed and unmanaged versions). With best regards, Andrew

[Lost User]

I guess "5" It's actually the same with C(++), because the expression i++ is evaluated AFTER the statement.

[Andrew Kirillov]

Greeeg wrote: It's actually the same with C(++), Not the same. I've checked on my VS 2003. With best regards, Andrew

[sreejith ss nair]

Problem is there in your example. Try to use different variable to distingush the difference. int i=5,j=0; j=i++; Sreejith Nair [ My Articles ]

[Andrew Kirillov]

Sreejith SS Nair wrote: Problem is there in your example. I know that my example is useless. It was interesting another thing: C(++) gives 6, but C# gives 5. Never mind, the C# behavior is good described in MSDN[^]. With best regards, Andrew

[sreejith ss nair]

You wrote : Provide me, any ideas why does C# compiler work so ? I post answer for your queary. And sorry if you twist the issue. Sreejith Nair [ My Articles ]

[leppie]

IMO this is a compiler bug. From the IL you can see there is some funniness. xacc-ide 0.0.15 now with C#, MSIL, C, XML, ASP.NET, Nemerle, MyXaml and HLSL coloring - Screenshots

[Dave Kreskowiak]

Andrew Kirillov wrote: int i = 5; i = i++; Easy, 5. The value of i is incremented after the expression is evaluated. In this example: Andrew Kirillov wrote: int i = 5; i = ++i; Here, i will be 6. The value of i is incremented before the expression is evaluated. This is all defined functionality in the C# language specification. C and C++ do the same thing. 7.5.9 Postfix increment and decrement operators[^] RageInTheMachine9532 "...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome

[Andrew Kirillov]

Dave Kreskowiak wrote: C and C++ do the same thing. No ! I wrote it alredy. I've tested it with VS.NET 2003. C++ (managed and unmanaged) gives 6. With best regards, Andrew

[Wjousts]

Andrew Kirillov wrote: No ! I wrote it alredy. I've tested it with VS.NET 2003. C++ (managed and unmanaged) gives 6. Then I'd say the problem is with C++ not with C#. Try this: int i = 5; if (i++ == 5) { Console::WriteLine("Five"); } else { Console::WriteLine("Six"); } In C# and C++ you should get "Five" as a result.

[Andrew Kirillov]

Wjousts wrote: Try this: .... In C# and C++ you should get "Five" as a result. Yes, I know. It will be the same in both languages. I was interested in original code I wrote. With best regards, Andrew

[Dave Kreskowiak]

So it does... It looks like your right. So what's the point of all this? RageInTheMachine9532 "...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome

[Wjousts]

Actually you are right, it should be 6 (in your original code) in both C++ and C#. Nasty trick question that. So it would appear to be a problem with the C# compiler.

[Dave Kreskowiak]

After doing some digging, the evaluation rules are slightly different between the C++ language specifications and the C# specs. It's not a bug as far the specifications describes the process. RageInTheMachine9532 "...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome

[Wjousts]

No, I think it is a bug, evaulation rules or not. It's very easy to miss (and I did at first) but the prefix/postfix thing is irrelevant: int i = 5; i = i++; versus int i = 5; i = ++i; Should both end up with i = 6. i++ is the same as i = i + 1 (so is ++i) so really the both code snippets should become int i = 5; i = i; i = i + 1; and int i = 5; i = i + 1; i = i; It shouldn't matter if you do i = i first or i = i + 1 first the result should still be six. As somebody else pointed out, it you do j = i++ then look at the value of j then the prefix/postfix part matters, but because you are putting the result back in to i it should be six either way.

[Dave Kreskowiak]

According the doc's I read in the specifications, the result your seeing is the expected result of the expression. C# in 2002, 2003, and 2005 do the exact same thing. The specifications between versions didn't change. RageInTheMachine9532 "...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome

[Wjousts]

The specifications concerning the behavor of postfix and prefix increment aren't relavent. The result should be 6 either way! i = i++; i gets assigned 5 and then i gets incremented. Therefore i = 6.

[Daniel Grunwald]

Wjousts wrote: i gets assigned 5 and then i gets incremented. Therefore i = 6. That's not the normal execution order. The real execution order (in C#) is: Read value of i Increment i Assign old value of i to i Why should the "++" wait until after the assignment? It directly binds to "i": AssignmentExpression: Left=i Op=Assign Right=[UnaryExpression: Op=PostIncrement Expr=i] It behaves differently in C++ because C++ has very strange rules for execution order.

[Dave Kreskowiak]

Wjousts wrote: The specifications concerning the behavor of postfix and prefix increment aren't relavent. WHAT?! :confused: Then why have a language specification at all? They detail the inner workings of the instruction set and execution order. According to the C/C++ language specifications, the value should be 6. Stop applying C/C++ specifications to C#. RageInTheMachine9532 "...a pungent, ghastly, stinky piece of cheese!" -- The Roaming Gnome