C# – Combine two integers to create a unique number

cintegermathstring

Good Morning,

I was looking for a way to combine two integers to create a unique number, I have two tables that I need to combine into a third table with unique numbers,

These are my tables:

Table A 
SchoolID    ClassId
107 56644231
107 56644532
107 320110212

Table B 
SchoolID    ClassId
108 566442310
108 56644532
108 50110212

I need to export these fields to a third table combining class ID and school ID into one single field called classID. I need to be able to combine these numbers together and then be able to uncombine them to get schoolid and classid separate for update purposes. I was thinking of concatenating the strings 'schoolid + '00' + 'classid' since I know that schoolid will always be a 3 digit number but I am looking for some other way perhaps mathematical where I don't have to use string casts.

Is there a mathematical way to do this? Or is casting to string the best way to do this?

I am using C# to code the solution.

Thanks,

Best Answer

Similar to Magnus Hoff, but I would recommend using a binary friendly approach instead of a base 10 approach.

combinedid = (classid << 8) + schoolid;

And then, later:

classid = combinedid >> 8;
schoolid = combinedid & 0xFF;

I think this is a little more straight forward from a programming standpoint (making it clear that your school ID is 1 byte (0-255), the class ID is 3 bytes).

You could also easily do this with a bigint (Long / Int64), making two int32's a single int64 safely:

combinedid = ((long)classid << 32) + schoolid;

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Python – Display number with leading zeros

In Python 2 (and Python 3) you can do:

print "%02d" % (1,)

Basically % is like printf or sprintf (see docs).

For Python 3.+, the same behavior can also be achieved with format:

print("{:02d}".format(1))

For Python 3.6+ the same behavior can be achieved with f-strings:

print(f"{1:02d}")

Determine Whether Two Date Ranges Overlap

(StartA <= EndB) and (EndA >= StartB)

Proof:
Let ConditionA Mean that DateRange A Completely After DateRange B

_                        |---- DateRange A ------|
|---Date Range B -----|                          _

(True if StartA > EndB)

Let ConditionB Mean that DateRange A is Completely Before DateRange B

|---- DateRange A -----|                        _ 
_                          |---Date Range B ----|

(True if EndA < StartB)

Then Overlap exists if Neither A Nor B is true -
(If one range is neither completely after the other,
nor completely before the other, then they must overlap.)

Now one of De Morgan's laws says that:

Not (A Or B) <=> Not A And Not B

Which translates to: (StartA <= EndB) and (EndA >= StartB)

NOTE: This includes conditions where the edges overlap exactly. If you wish to exclude that,
change the >= operators to >, and <= to <

NOTE2. Thanks to @Baodad, see this blog, the actual overlap is least of:
{ endA-startA, endA - startB, endB-startA, endB - startB }

(StartA <= EndB) and (EndA >= StartB) (StartA <= EndB) and (StartB <= EndA)

NOTE3. Thanks to @tomosius, a shorter version reads:
DateRangesOverlap = max(start1, start2) < min(end1, end2)
This is actually a syntactical shortcut for what is a longer implementation, which includes extra checks to verify that the start dates are on or before the endDates. Deriving this from above:

If start and end dates can be out of order, i.e., if it is possible that startA > endA or startB > endB, then you also have to check that they are in order, so that means you have to add two additional validity rules:
(StartA <= EndB) and (StartB <= EndA) and (StartA <= EndA) and (StartB <= EndB) or:
(StartA <= EndB) and (StartA <= EndA) and (StartB <= EndA) and (StartB <= EndB) or,
(StartA <= Min(EndA, EndB) and (StartB <= Min(EndA, EndB)) or:
(Max(StartA, StartB) <= Min(EndA, EndB)

But to implement Min() and Max(), you have to code, (using C ternary for terseness),:
(StartA > StartB? Start A: StartB) <= (EndA < EndB? EndA: EndB)

Best Answer

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