Comments by programmer095 (Top 28 by date)

Finally! Understand. Accepted.

A 5. Enough for practically all purposes.

Thank you so much. I accept answer.

Thank you so much. I accept answer.

So I did, thanks. Probably I smile along. How about you? ;-)
Many thanks for your Real Help, jerrykid.

Oh, that's great!
I think I just need to think. You gave me enough food for though for now.

Thank you so much!

Surprised. This is just the opposite to my explanation. I though it will not compile and I did not understood why my variant did not compile.
I still don't understand.

Anyway, I'll think if I can use it. Thank you very much. It would be good to undestand why.
You rock! (unlike others)

I understand and apologize. I value the help very much.

I don't understand only one thing: if I call lie a truth, I will merely get yet another lie.

Well, sorry. I did not want to hurt your feelings. Thank you for advice.
I already got very good help.
And I did not give good answers because I am not as strong as you are.

What attitude do you mean? Did I tell you anything which is not true? I did not ask you personally to help, I hoped for real help. If nobody helps, fine; what's not fair?
As I have so poor skills, it let you fool me with your answer. It was easy but is it fair? "This guy knows very little, why not feeding some crap to him?", is this a way?

It just happened I tried to compile it. Actually, you gave me much better lesson: not to trust anyone, even so knowledgeable as you are. Thank you very much.

It's even worse: Error 1 'CompilationEventArgs.implicit operator UserActionEventArgs(CompilationEventArgs)': user-defined conversions to or from a base class are not allowed

Sure. It is not allowed because it is already implicitely assignable.

Excuse me, why whould you answer with what you did not try? Thank you, but not answering is better.

Thank you, but it will not help.
This is my point: UserActionEventArgs is the base class for CompilationEventArgs, not the other way around. It means it is already assignable the way you put it. Still, it does not compile.

Did you compile it? I'm afraid to try. I don't want to write another comment like I did for "answer" by Delicious Bisquit.

I see, thank you. My variable in real not int, it is double

Who will need to check infinity after every division?! With C++ there was exception -- all I need

Ok, I just tried it. It did not compile.

First, what's the difference? I explained why I don't understand why.

Second: why would you lie to me?!
I have a code with 12 lines only. After your advice I had to modify only one line.
Before answering you could type it and try.
I saw your answer will not work before I tried, because there is not difference. But I tried.
Thanks for try. But you know, lie is much worse then saying nothing.
I vote big fat 1.

I don't know. Why mode code is not working?

Yes, of course! I've tried this recently before looking at your answer. But the answer is correct. This is simple. The manual of the board is hard to understand. Of course I will accept the answer. Thank you.
I still think this two's complement is weird. Flipping bit is just one bit operation. Comparing two absolute value is easy, just compare all buts except sign bit. Why flipping so many bits for just a negation operation? Who designed such thing? Are all the processors so weird? What DAC does is right, what processor does is just to confuse me, what else?

I guess it should work. I guess Clipboard really was misleading. My 5

I've read the ECMA standard, thank you. Look at first sentence in 8.11.1: "Method definitions are composed of a name, a method signature, and optionally an implementation of the method." But is there is no implementation, only keyword "abstract" can make it a definition, see below: "A method definition that does not include a method implementation shall be marked as abstract.".

I did not mark MethodC abstract in my code, so it means first line with the method declaration cannot be a method definition. So, this is a declaration, not definition. 3 lines below there is a definition because there is an implementation. So, "alredy defines" error message cannot refer to this definition in first line. Is should be something else. What?

Thank you very much. I'm looking at this article: is it about some real machine? And why they do such complex calculations? just to make CPU working slowly? And I still does not understand how this trick is related to my question, sorry...

Sorry, I just tested your example on two integer types: byte and unsigned byte. In both cases:
100 = 4
101 = 5
110 = 6
111 = 7
Besides, how can possibly it make any sense: a jump from 3 to -4. Why?!

Ok, but I still cannot understand the error message. : 'Program' already defines a member called 'MethodC' with the same parameter types. "MethodC()" is not a method definition, only "MethodC() {}" is. So, there is no double definition. How just the declaration of a method can prevent compilation?!

I cannot understand this. How can it possibly work? When MethodB is called, it is not defined.