Looking for a shorter way of equating

[sayzer]

Hi there,
I am wondering if there is a much shorter way of doing this?

'VARX is a word.

VARX = 56789 
'I need digit4 and digit3 (56) from this number in to a byte variable; 

'I do it in this way:
VAR_Fisrt = (VARX DIG 4)*10 + (VARX DIG 3)
'This gives me 56.

'I also need digit2 and digit1 (78) in to another byte variable;
'I do it in this way:
VAR_Second = (VARX  DIG 2)*10 + (VARX DIG 1)
' This gives me 78.

Thanks inn advance.

[Darrel Taylor]

sayzer,

Don't know if it's any better or not, but here's a different way.

VARX = 56789
'I need digit4 and digit3 (56) from this number in to a byte variable;

'I do it in this way:
VAR_Fisrt = (VARX DIG 4)*10 + (VARX DIG 3)
'This gives me 56.

VAR_Fisrt = VARX / 1000

'I also need digit2 and digit1 (78) in to another byte variable;
'I do it in this way:
VAR_Second = (VARX DIG 2)*10 + (VARX DIG 1)
' This gives me 78.

VAR_Second = (VARX / 10) // 100
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[sayzer]

Thanks DT, but what about the rounding?

I thought that it is going to round the result when divided by 1000.
In our case, we should be getting 56 dividing by 1000, but since the next digit is 7 which is over 5, the actual result will be 57 (56 will be rounded to 57)

Or is it not?

[Darrel Taylor]

Nope,

No rounding involved. It only gives the integer value of the division.

DT