Strange IL code from simple C# expression
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
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 -
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..) SiAnonymous 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
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
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I guess "5" It's actually the same with C(++), because the expression i++ is evaluated AFTER the statement.
Greeeg wrote: It's actually the same with C(++), Not the same. I've checked on my VS 2003. With best regards, Andrew
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
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 ] -
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 ] -
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 ]
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
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
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Hello, guys I found an interesting topic on MS newsgroup. And I'd like to ask you a simple question. What is the value of "
i" variable after the code ?int i = 5; i = i++;??? Provide me, please, with an answer you though/expected just after you looked at the code. And then, any ideas why does C# compiler work so ? (if you will run the code, you will understand what I mean). With best regards, Andrew
Andrew Kirillov wrote: int i = 5; i = i++; Easy, 5. The value of
iis incremented after the expression is evaluated. In this example: Andrew Kirillov wrote: int i = 5; i = ++i; Here, i will be 6. The value ofiis 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 wrote: int i = 5; i = i++; Easy, 5. The value of
iis incremented after the expression is evaluated. In this example: Andrew Kirillov wrote: int i = 5; i = ++i; Here, i will be 6. The value ofiis 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 GnomeDave 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
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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
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 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.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
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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
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
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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
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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.
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
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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
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++;versusint 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 becomeint i = 5; i = i; i = i + 1;andint i = 5; i = i + 1; i = i;It shouldn't matter if you doi = ifirst ori = i + 1first the result should still be six. As somebody else pointed out, it you doj = 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. -
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++;versusint 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 becomeint i = 5; i = i; i = i + 1;andint i = 5; i = i + 1; i = i;It shouldn't matter if you doi = ifirst ori = i + 1first the result should still be six. As somebody else pointed out, it you doj = 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.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
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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
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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.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.