I need to know how to "split" a number up into seperae numbers, using VB. For example, "54321". I need to split that into 5 seperate numbers - "5," "4," "3," "2," "1," and assign them to a variable, if possible.

This next part isn't important, but if you're curious why I need to know ...
Eventually what I need to do is split the number into these individual parts and add them together. With that new number I need to again split it and take the last number subtracted from 10. That new number will then be placed at the end of the beginning number.

Example:
5+4+3+2+1 = 15
15 = 1 & 5, take the last number ..
10-5 = 5
New number = "543215"

The thing is I have no clue how to split this number into the seperate parts. Once I get past that I'm 99% sure I can wrap up. Anyone able to help? If so, Thanks!

This post has been edited by miCRoSCoPiC^eaRthLinG: Dec 30 2005, 03:48 AM

That should be fairly easy with the String manipulation routines that accompany VB. There are infact two methods you can follow.

Method 1:
    [/tab]To start with, you can simple convert the number into a string - and then use the substring function to find out each character. To achieve this we'll have to find the length of the string - which is 5, in your example above - and then iterate through each position of the string, picking up individual characters as we go.


That's it - the above approach will give you an array called Digits() whose positions indexed by 0 through Length - 1 will contain the individual digits of your number.

[hr=noshade][/hr]
Method 2:
[tab]The second method essentially involves the same principle as above - but computationally is a little faster - as the algorithm is based on numerical computations. A few words on the algorithm first.

    [/tab]Take for example, your own number, 54321 - can you think of a concrete method of stripping the number off the first digit "5" - leaving behind only "4321" ??
What would give you 5 as a result, if you divide 54321 by it ? You'd instantly arrive on the figure 10,000.
So: 54321 / 10000 = 5
This is all fine - you've got your first digit. But then how do you get the rest of the number together ? Arithmatically, if you divide 54321 by 10000 - we'd have a remainder of 4321 - that's exactly what we need. So how do you find the remainder. Here's where an arithmatic operator called Mod comes into play. It's known as Mod in terms of Visual Basic - and mostly represented as a % symbol in languages like C/C++, Java etc.
So: 54321 Mod 10000 = 4321

There - we have our first digit extracted, leaving behind the last four. Now if you repeat the above steps, except that you divide and do the Mod with 1,000 instead of 10,000 - you'd yet extract one more digit leaving behind the last 3. Next you repeat the steps again with 100 and then with 10 - and you'd have all your digits separated out.

How do we implement this ? Here's the code:


Even in this case, your array Digits() will contain the extracted digits starting from position 0 through Length - 1.

[tab]Try out both and let me know - both should work just fine, giving your the same output. Personally, I find the second method more ELEGANT. The first one is a little blunt - doing things by brute force, although is much easier to understand. However, as I mentioned earlier, the second one computes way faster.

All the best
m^e

Well it would be relevant if you posted which language you wanted this solution for. I think converting a number into string and then converting into numbers again is an easy solution. In C++ probably the shift operatros would do the trick.

Oh yeah - that reminds me, since you made the post in VB.NET forum - the solution provided was in that same language too.

Also hatim - if you notice the solution I provided in Method 1 does exactly the same - conversion to string, extraction and then back to integer - and the second method entirely eliminates the need to do that, thus making the algorithm far more efficient.

Since I hate to leave people hanging ..

I went with the second method, and it works excellent! Much thanks.

Awesome. Glad that helped That second method was a chance discovery while I was experimenting with coding a bunch of numerical manipulations quite sometime back - in relation to this cryptographic routine that I was trying to come up with.