Why this method does not increment?

Question

2.14/5 (5 votes)

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May be I had to much drinks last night. Why this method does not increment the value of "number" by 1 every time I receive the value "1" in "getTurns"?

class Turns
    {
        private int number;

        public void Counts(int getTurn, out int getBack)
        {
            if (getTurn.Equals(1))
            {
                number++;
            }
           
            getBack = number;

        }
    }

Posted 26-May-13 5:43am

Millone

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Comments

Richard MacCutchan 26-May-13 12:04pm

It works fine for me; show how you are calling it.

Sergey Alexandrovich Kryukov 26-May-13 12:07pm

Not
if (getTurn.Equals(1)) {...}
but
if (getTurn == 1) ...!

And there is need to "out" as you can do return number.

If you are going to write is such inaccurate way, feeling drunk from time to time will be guaranteed!

—SA

Millone 26-May-13 12:36pm

@ Sergey. I asked for support and it was not my intention to confront, but because you are asking, I reply. I am using out because I need to add other parameters and "return" would not allow me to send back more than one variable. I could return an array but "out" is fine enough to send back 2 variables. Concerning "Equals" does exactly the same of "if" in this case, so I do not see the point. I have sorted out the problem that was of a completely different nature.

Sergey Alexandrovich Kryukov 26-May-13 13:36pm

What confront?! You are getting an advice, because you really need it, and you are talking about confrontation. You are not trying to be constructive. No, you can use return as well, only it would be more neat. Equals are not exactly the same: it gives the same result with lower performance.

Look, you are lost, you are very much lost. You really need help, but we hardly can provide you this kind of help. You simply have to forget all you "know" about programming and start from the very beginning. Take a manual and read, doing simple exercises. Only in this way you get get out of the dead and you are in.

—SA

Manfred Rudolf Bihy 26-May-13 12:14pm

Should work fine, except that I'm SA on this one. A completely other thing is making this method void. Why not just return the incremented value? There should be an example of your usage pattern to explain that.

Ian A Davidson 26-May-13 12:41pm

I'm with Richard. Show us how to you are calling this.
Ian.

Millone 26-May-13 12:45pm

Ian, it is working now. See my solution above. The problem was that I was basically calling the Count method from two different instances of the Turns class. When number was an instance field, it would start off at 0 for each instance. However, when I changed it to static, it would start off at 0 for the class as a whole and would be incremented by 1 as each instance called the Count method.

4 solutions

Solution 4

@Millone, please, read comments!

Manfred R. Bihy is right that you should care about instance field initialization. Please, read these articles:
10.4.5 Variable initializers[^]
10.4.5.1 Static field initialization[^]
10.4.5.2 Instance field initialization[^]
Instance Constructors (C# Programming Guide)[^]
Static Constructors (C# Programming Guide)[^]
Static Classes and Static Class Members (C# Programming Guide)[^]
Access Modifiers (C# Programming Guide)[^]

Millone wrote:
The problem was that I was basically calling the Count method from two different instances of the Turns class. When number was an instance field, it would start off at 0 for each instance. However, when I changed it to static, it would start off at 0 for the class as a whole and would be incremented by 1 as each instance called the Count method.

Probably, you don't need to call another instance of your class, but you don't know how to access to existing one...
3.7 Scopes (C#)[^]

I would suggest you to read about delegates[^] and interfaces[^].
Using delegates or/and interfaces you'll be able to increase Counts value as many times as you wish, using only the one instance of your class.

Bad practice is always bad practice, no matter of result (working application)!

Posted 26-May-13 10:42am

Maciej Los

Updated 27-May-13 1:52am

v3

Comments

Ian A Davidson 26-May-13 17:24pm

While some of what you have here is generally good, this is not addressing the question at all. We don't know the rest of the code, or what he is trying to do.

Maciej Los 26-May-13 17:44pm

Thank you for your comment, but i can't agree with you at all (except the first part of your comment: "generally good"). When OP will read linked articles, he'll understand his mistake.

1 star for "generally good" answer? Is it justifiable?

;)

Ian A Davidson 26-May-13 18:03pm

Sorry you disagree. I was being polite. Some of the advice given would be good if the question was different. (Although, why you would bother with "number" at all in this case I'm not sure; I'd go with "public int Count { get; set; }" ) However, as the question stands, there is nothing here to help. Hence 1 star.
Ian.

Maciej Los 26-May-13 19:31pm

;(

Solution 2

~~How does this even complile? The variable number of type int has never been initialized.~~
I stand humbled and corrected by the comments below, but I still think that uninitialized value and to some lesser degree also reference types in class members are an absolute no go. I always assign a value even if it may seem a bit verbose, not even the beginner is left guessing what is actually going on.

My two cents!

Regards,
Manfred

Posted 26-May-13 6:18am

Manfred Rudolf Bihy

Updated 26-May-13 8:44am

v2

Comments

Ian A Davidson 26-May-13 12:35pm

What do you mean? I can't see anything wrong with this.
Ian.

Millone 26-May-13 12:46pm

@Manfred, it did!

Sergey Alexandrovich Kryukov 26-May-13 13:39pm

Manfred, sorry, but you did one mistake: this is not a variable, this is a class member, it does not require initializer, because it is initialized by default. Remember the syntax (or just try it out).

I would advise you to remove this incorrect post.
—SA

Manfred Rudolf Bihy 26-May-13 14:31pm

Thanks for pointing that out SA, I'll amend my post.

Ian A Davidson 26-May-13 16:24pm

Manfred, your comment is still incorrect. In C# (or any other .NET language, i.e. if you're using the CLR), initialising a field (either static or instance field) of a class to its default value is redundant (except in some rare cases), and so you're just adding a performance hit to your application (admittedly probably only a problem if you are creating thousands of those objects, but it is good practice not to do it). Read here: http://msdn.microsoft.com/en-US/library/ms182274(v=vs.80).aspx
Regards,
Ian.

Manfred Rudolf Bihy 26-May-13 16:37pm

My expression in the above statement is one of personal preference of clarity above brevity. I'm curious how this can be "incorrect". Feel free to share your thoughts though.

Ian A Davidson 26-May-13 17:06pm

You solution says "uninitialized value ... [is] an absolute no go." If this was C++ I would agree with you, but this is C# so fields are never uninitialized, and the compiler is intelligent enough to tell you whenever you attempt to use a value that is not initialised, and won't compile in that case. Integers are initialised to zero (and, by the way, reference types are initialised to null).

Manfred Rudolf Bihy 26-May-13 17:42pm

As I stated above: clarity above brevity. Regarding your post about my comment to optimizations: If it is bad practice to always initialize variables and class members, then I consider myself guilty. The comment below on relying on optimizations is true though, even when you meant it in a negative, polarizing way. I've already two written compilers and truely believe that optimizing code generation is the way to go. Anyhow it's always great to hear some other opinion. Otherwise life would get too boring.

Ian A Davidson 26-May-13 18:01pm

I didn't say "it is bad practice to always initialise variables and class members". It IS bad practice to initialise fields ("class members") to their default value, since they are initialised to that already, so you're doing it twice. If you want clarity, put a comment to state that the default value is what you want it set to. I.e. in the above case, "int number; // number initialised to zero". I don't believe this is a "clarity / brevity" issue, but a pure misunderstanding of the C# language and/or ethos. As to variables, you always need to initialise them, and the compiler will complain if you try to use one before it's been assigned to. All I can say is you have obviously never written a compiler for the CLR (at least, I hope not...)
Regards,
Ian.

Manfred Rudolf Bihy 26-May-13 16:41pm

Besides that, any compiler worth it's money can always optimize away initialisations to the default value. Still you made me curious, so I'll read up on that link you posted. Thanks for the information!

Ian A Davidson 26-May-13 17:08pm

I love the way so many people seem to think that the compiler can/will "optimise" almost anything. It seems to be an excuse for many a bad piece of code and bad practice.
No offence intended, just trying to help people learn here, and that sometimes means having to refute mis-information.
Regards,
Ian.

Manfred Rudolf Bihy 26-May-13 17:44pm

Initialization a bad pracice ...
Hmmmmm, see my comment above.

Maciej Los 27-May-13 7:11am

I agree with Manfred in part where "compiler ... always optimize away initialisations to the default value." This article is a proof: Variable initializers C#[^]

Ian A Davidson 27-May-13 11:43am

I don't see it. The link you have given seems to show the exact opposite. Firstly no where on the page is the default value assigned. The default assignment is not optimised away! In fact the example at the bottom of the page (which as it says is strongly discouraged), relies on the very fact that it is not optimised out - the fields have a default value before the assignment initialisation runs, otherwise the code wouldn't work: a is initialised to 1, because a and b are first assigned their default values of zero, before the initialisers of the class run, so when a's initialiser runs a = b+1 = 0+1 = 1, and then b's initialiser runs, b = a+1 = 1+1 = 2.
Regards,
Ian.

Maciej Los 27-May-13 7:13am

+5!
It's very important to feel the difference between good and bad programing practice.

Manfred Rudolf Bihy 27-May-13 7:25am

Thank you, at least somebody sees it my way! :)
I've never seen this many downvotes on one of my posts before. I wonder what kind of "think" is going on here.
I'll stand by my conviction though.

Cheers!

Maciej Los 27-May-13 7:35am

I'm with you, Manfred! Head up!
You'll see my answer soon...

Maciej Los 27-May-13 9:04am

Please, see my answer. Now it has been restored.

Simon_Whale 28-May-13 7:07am

+5 from me Manfred, I do this approach all the time, for me I personally use it to help me debug issues as if it has a default value I know when a piece of code has worked or not.

Solution 1

try this

C#

class Turns
{
public object Counts(int count)
   { 
	if (count == 1) 
        {
	   count += 1;
	}
	return count;
   }
}

Posted 26-May-13 6:17am

Basmeh Awad

Updated 26-May-13 20:28pm

v2

Comments

Bojjaiah 27-May-13 1:01am

Hi basmeh,
void type doesn't have return type?

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Millone · Accepted Answer · 2013-05-26T06:29:00

I changed the "private int number" to "static int number" and its working fine.
The problem was that I was basically calling the Count method from two different instances of the Turns class. When number was an instance field, it would start off at 0 for each instance. However, when I changed it to static, it would start off at 0 for the class as a whole and would be incremented by 1 as each instance called the Count method.

class Turns
    {
        static int number;
 
        public void Counts(int getTurn, out int getBack)
        {
            if (getTurn.Equals(1))
            {
                number++;
            }
           
            getBack = number;
 
        }
    }

Why this method does not increment?

4 solutions

Solution 4

Solution 2

Solution 1

Solution 3

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