uniqueidentifier

Possible Duplicate: Python “is” operator behaves unexpectedly with integers I stumbled upon the following Python weirdity: >>> two = 2 >>> ii = 2 >>> id(two) == id(ii) True >>> [id(i) for i in [42,42,42,42]] [10084276, 10084276, 10084276, 10084276] >>> help(id) Help on built-in function id in module __builtin__: id(...) id(object) -> integer Return the identity of an object. This is guaranteed to be unique among simultaneously existing objects. (Hint: it's the object's memory address.) Is every number a unique object? Are different variables holding the same elemental values (for example, two

Consider I've two numbers 1023232 & 44. I want to generate a unique number representing this combination of numbers. How can i generate it? Requirement f(x,y) = f(y,x) and f(x,y) is unique for every (x,y) or (y,x) if those are two ints, you could just do this: ulong F(int x, int y) { ulong id = x > y ? (uint)y | ((ulong)x << 32) : (uint)x | ((ulong)y << 32); return id; } if you need to generate a truly unique value for two variables of a given size, you need about double the size of each variable. (ok, a bit less now that f(x,y) == f(y,x)) You could also get your original values back by

This question already has an answer here: Unique Identification of iOS device for iOS 7.0 and above 6 answers In my iOS app, I have to restrict the user to use iOS app per device. To do this I found a solution that we can use the UUID (Universally Unique Identifier) or UDID (Unique Device Identifier) . But according to this answer I can't use UUID, because if app gets deleted or reinstalled UUID has been getting changed and I don't want this. Also Apple rejects apps if app uses UDID . Is there any way to identify iOS device uniquely. Apple has done away with the approach of UDIDs and will

Possible Duplicate: UIDevice uniqueIdentifier Deprecated - What To Do Now? Even if Apple was not at Barcelone's MWC (mobile world congress), there was the certitude that getting the deviceID will be deprecated in further iOS SDK. I do not understand why Apple want to restrict this, but that's not the topic. I must prepare my application to an alternative because my users are identified and known for a better use of my app (don't need to log, or create an account, for example). And I'm sure I'm not alone in that case. So anybody know an alternative from getting the deviceID ? Is there other

On an iPhone I can use [[UIDevice currentDevice] uniqueIdentifier]; to get a string which identifies this device. Is there anything equal in OSX ? I didn't find anything. I just want to identify the Mac which started the application. Can you help me ? Apple has a technote on uniquely identifying a mac. Here's a loosely modified version of the code Apple has posted in that technote... don't forget to link your project against IOKit.framework in order to build this: #import <IOKit/IOKitLib.h> - (NSString *)serialNumber { io_service_t platformExpert = IOServiceGetMatchingService

Basically I want to use uniqueidentifier in similar way as identity. I don't want to insert values into it, It should just insert values automatically, different value for each row. I'm not able to set autoincrement on columns of type uniqueidentifier(the property 'autoincrement' is set to false and is not editable). Or even better: use the newsequentialid() as the default for your UNIQUEIDENITIFER column. That'll give you a somewhat sequential series of GUIDs. CREATE TABLE dbo.YourTable (SerialID UNIQUEIDENTIFIER CONSTRAINT DF_SerialID DEFAULT newsequentialid(), .... (other columns)...... )

Objective: I am looking for a way to find out a unique_id for android device. Background: I will use the Id in login request payload and as my app is license based service app the Id should not change under normal circumstances. Existing Approaches: In iOS there are some unique id solutions for iOS such as CFUUID or identifierForVendor coupled with Keychain ,Advertising Identifier etc.. that can do this job upto the expectation. But in Android all the options that I know seems to have hole in it. IMEI: TelephonyManager TelephonyMgr = (TelephonyManager)getSystemService(TELEPHONY_SERVICE);