Re: Shift operator - C# Discussion Boards

18-Dec-09 3:23

I'd like to convert an integer to a byte. Then going through some process steps (i don't know how to do and need your help), finally when apply (b<<24) i'd like it return to the integer's value before. How I can do that?

int temp; // temp has a arbitrary value
byte b;
// Processing steps with b
//...
//...
// to finally when apply (b<<24)
int result=(b<<24);//expected value equal to temp.

Thank you so much!

modified on Friday, December 18, 2009 10:55 AM

thungphan18-Dec-09 4:30

18-Dec-09 4:30

I did following but result is 0, not 5.

int temp=5;
byte b=(byte)(temp>>24);
int result=b<<24;
Console.WriteLine(result);

Could you post some code? Thanks your reply!

Rob Philpott18-Dec-09 4:42

Rob Philpott

18-Dec-09 4:42

That's because your shifting all your bits to the right and off a cliff.

int temp = 5;
int t1 = temp << 24;
int t2 = t1 >> 24;
Console.WriteLine(t2);

Regards,
Rob Philpott.

thungphan18-Dec-09 4:54

18-Dec-09 4:54

I'm sorry. My question is not correct. I modified it. Please read it again and help me!.
Thanks so much!

harold aptroot18-Dec-09 4:55

harold aptroot

18-Dec-09 4:55

What you want is in conflict with what you want. For starters there is no way to convert "any int" to a byte (proof: there are more possible values for an int than for a byte)

What are you trying to do with this calculation? Maybe we can suggest an alternative?

Luc Pattyn18-Dec-09 5:09

Luc Pattyn

18-Dec-09 5:09

Hi,

you probably are confusing two things:

1.
assuming b is unsigned or positive, b<<24 is an attempt to generate a larger number, in fact it is equivalent to a multiplication by 2^24 (which might overflow, i.e. loose bits shifted out at the high end, and therefore could result in zero). Anyway, it is a numeric or mathematical operation. So is 5>>24 which will result in zero since all bits get lost at the low end, as others have said already.

2.
multi-byte variables (such as int) are stored in memory in one of two ways:
"big endian" = the most significant byte comes first (i.e. at the lowest address)
"little endian" = the least significant byte comes first.
Intel processors (and some others too) implement the little-endian convention. Other processor families, and some networks and protocols use big-endian (e.g. the administrative information used on Ehternet networks).

Both issues are unrelated. If you want to operate on values/numbers, look at (1). If you want to operate on memory bytes, don't look at (1), use the BitConverter class instead (or a fake "union" construction, with a struct and explicit offsets).

Smile | :)

Luc Pattyn [Forum Guidelines] [Why QA sucks] [My Articles]

I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages

modified on Friday, December 18, 2009 11:44 AM

thungphan18-Dec-09 17:05

18-Dec-09 17:05

byte b0=GetByte(); // after these four commands
byte b1=GetByte(); // b0, b1, b2, b3 have values
byte b2=GetByte(); // 
byte b3=GetByte(); //
int vr= + b1;
if ((b0<<8+b1)==0x4F42)
{
    int length;
    if (littleEndian)
       length=(b3<<24) + (b2<<16) + (b1<<8) + b0;
    else
       length=(b0<<24) + (b1<<16) + (b2<<8) + b3;
}

Now I have a value of length variable and to convert inverse direction. I'd like to gain b0, b1, b2, b3.
And those gained b0, b1 must ensure when apply (b0<<8) + b1 will be equal to 0x4F42 again.
That's what I want to. Before i try to simplify my problem but unfortunately i made it difficult to understand for you.
Please help me! Thank you so much!

Hide database file from user

harold aptroot19-Dec-09 3:29

harold aptroot

19-Dec-09 3:29

Well if that's all then we gave you the answer countless times, right?

Two ListViews

Tagamoga18-Dec-09 2:38

Tagamoga

18-Dec-09 2:38

Hello.

I have got two ListViews in one Form.

When I select one DataSet in the first one, the second one will be filled with the correct data. Now I shall allow to select a dataset in the second one. But, when I do this, the selection in the first one is canceld. How can I awoid this deselection? Both seletions should be shown.

Regards, Taggi

How do you know, that Tagamoga has been assimilated by the borg?
"Recistunce ist fiutille."
Sorry for violating the english language.

Re: Two ListViews

benjymous18-Dec-09 3:14

benjymous

18-Dec-09 3:14

There's a property on the listview - I can't remember what it's called, but it's something like "Always show selection". The default listview behaviour is to hide the selection highlight when the control loses focus (which is what happens when you click in the second control)

Help me! I'm turning into a grapefruit!
Buzzwords!

Re: Two ListViews

Tagamoga18-Dec-09 3:29

Tagamoga

18-Dec-09 3:29

Thank you very much!

The Property is called "HideSelection" and has to be set on "false".

Thanks!

How do you know, that Tagamoga has been assimilated by the borg?
"Recistunce ist fiutille."
Sorry for violating the english language.

zedgraph

jashimu18-Dec-09 2:16

jashimu

18-Dec-09 2:16

Hi All,
Is anyone happen to know zedgraph. I am working on it for a while but can not figure out how to set X-axis value as hour.

thanks.

Message Closed

18-Dec-09 2:50

stancrm

18-Dec-09 2:50

Message Closed

modified 23-Nov-14 7:16am.

Re: zedgraph

jashimu18-Dec-09 4:09

jashimu

18-Dec-09 4:09

Hi stancrm, thanks for your help.
I have tried all of this but did not what I want.
what I want is that xaxis should show entire day as hour. let say
if from 12am to 11 pm and each minortic or majortic should show individual hour.
this what I am doing. I also know that xaxis depends on x value.If you can give me a hint
how do I frame time for the entire day as hour.

code>// Set the XAxis to date type
myPane.XAxis.Type = AxisType.Date;
// X axis step size is 1 day
myPane.XAxis.Scale.MajorStep = 1;
myPane.XAxis.Scale.MajorUnit = DateUnit.Hour;
myPane.XAxis.Scale.Format = "h tt";
myPane.XAxis.Scale.Min = hList[0].X - 1;

thanks again.

NarendraSinghJTV18-Dec-09 1:53

18-Dec-09 1:53

codeproject helped me many times
and i m thankful for that.

There is another question
when we create setup for windows application. can we set any property to hide the application database from the user. I want to hide the mdf file from the user.

thanks for all the reply in advance

Regards
Narendra Singh
(Jindal Tech Ventures)

Ashfield18-Dec-09 2:02

Ashfield

18-Dec-09 2:02

No more than you can hide any other file - you could set the hidden attribute, but thats about all. Why not use encryption within your database?

Bob
Ashfield Consultants Ltd
Proud to be a 2009 Code Project MVP

NarendraSinghJTV18-Dec-09 2:33

18-Dec-09 2:33

can we define the hidden property at the time of creating setup project and how can i use the encryption in database
if u please define some more it will be very helpfull for me

Regards
Narendra Singh
(Jindal Tech Ventures)

Saksida Bojan18-Dec-09 3:57

Saksida Bojan

18-Dec-09 3:57

The hidden attribute is a file atribute. It hides files from view for simple users. However more advance users uses 3rd party file manager or uses explorer setting "Show All Files". That settings shows files with Attribute: Hidden and System atribute.

One whay to fool users is to use .mdf database with custom extension (something.Dat).

If you realy want to prevent users from using your database, use on the fly encryption/decryption. I do not recommend to use hide attribute.
You could use windows folder and put somewhere or Use custom partition and use it to store database there. The second option is not recommended, because it is too hard and too long to successfully use it.

NarendraSinghJTV18-Dec-09 17:51

18-Dec-09 17:51

thanks saksida
now ill try to do encryption on database
can u provide me some links for startup

Regards
Narendra Singh
(Jindal Tech Ventures)

Simple encrypting and decrypting data in C#[^]

Saksida Bojan18-Dec-09 20:42

Saksida Bojan

18-Dec-09 20:42

Ashfield18-Dec-09 8:53

Ashfield

18-Dec-09 8:53

In reality there is no real way to hide the file, anyone with a bit of knowledge will ab able to find the file whatever you do. The only way to be secure is to use sql server encrytion, so just google that.

Bob
Ashfield Consultants Ltd
Proud to be a 2009 Code Project MVP

NarendraSinghJTV18-Dec-09 17:50