Really? I got it to work with multidigit numbers
Using:

Text(25,25,"0",A->Dec

Also a somewhat relating question, how would you put together 2 strings and store it into a new string?

Thanks that works. I just would wanna know if i could save "05" to something. Like this:
"0",A>Dec->B

You're my savior man. I'll try it soon.

EDIT: Works brilliant.

Wouldnt make sense to have any

Quote from: epic7 on November 22, 2011, 10:06:42 pm

Wouldnt make sense to have any

..unless it's part of a line of code with variable inputs, set up in a way that might have it happen.

Exactly. I may use something like that myself. So will that copy itself to itself?

Indeed, nothing bad would happen. It would take just as long as otherwise, and do nothing.
One thing to be aware of, however, if you are copying to/from a variable location, is to deal with overlap properly.
For example, it should be easy to see that this code will not shift 100 bytes, but will copy one byte 100 times:
Copy(L₁,L₁+1,100)

It's not always the same though.  If you Copy(L1,L1+1,100), it will first copy L1 into L1+1, and then L1+1 into L1+2, effectively filling L1 to L1+100 with the first element, instead of shifting. 

Quote from: Builderboy on November 22, 2011, 10:35:06 pm

It's not always the same though.  If you Copy(L1,L1+1,100), it will first copy L1 into L1+1, and then L1+1 into L1+2, effectively filling L1 to L1+100 with the first element, instead of shifting. 

To fix:
Copy(L1+98,L1+99,100){r}
(I think)

You mean Copy(L1+99,L1+100,100)

That's right. But the same thing happens when you use a For() loop—look at this.

As you said, this:

Copy(L1,L1+1,10)

does essentially the same as this:

For(A,0,9)
{A+L1}→{A+L1+1}
End

But if you take it step by step, it's really doing this:

{L1}→{L1+1}
{L1+1}→{L1+2}
{L1+2}→{L1+3}
{L1+3}→{L1+4}
{L1+4}→{L1+5}
{L1+5}→{L1+6}
{L1+6}→{L1+7}
{L1+7}→{L1+8}
{L1+8}→{L1+9}
{L1+9}→{L1+10}

See the problem? You're copying the first byte to the second, then the second byte to the third, and so on, so in the end you're just copying the first byte to every byte after it for ten bytes. It would be doing the same as this:

Fill(L1,10)

which isn't what we're trying to do.

If you want to just shift the whole block to the right a few bytes, you start at the end and use Copy()^r. Just thought I'd make that a bit more clear