How can I get 4.17 from 4.16666667

[daved]

x = 50/12

print x 'result is 4.16666667

x$ = str$(x)

print using("##.##",x$) 'result is 4.17

I need x to = 4.17 for a math calculation, but all I do results in 4.16666667
The print using("##.##",x$) result is 4.17
I've tried converting x to a string, then converting x to 4.16 (2 characters after the decimal),
and adding .01 if the last character is > 5. But when I convert back to an integer,
the result is 4.17666667, which is too large for an accurate calculation. 
How can I get 4.17?

[Walt Decker]

Try this:

X = 50 / 12
Res$ = STR$(VAL(USING("##.##", X)))

PRINT Res$

X = VAL(Res$)

[daved]

Perfect - That works!!

Thanks!!!

[tsh73]

a function that rounds to given number of decimals
for n = -5 to 5
print "Rounds to ";(n);" decimal places"
for i = 1 to 4
   rand=rnd(0)*1e6
   x = round(rand, n)
   print rand, x, using ( "#########.######", x)  'without USING you'll have only 7 digits printed
next
next

function round(x, n)    'rounds x to n-th digith (involves 1/10^n, obviously
   round=int(x*10^n+.5)/10^n  '+0.5 for rounding to nearest
end function



[honky]

If the third decimal is greater than 5, it is normal to round the second to the next higher unit.

[dan1101]

If you are just trying to kill the repeating decimal, you could cut it off at three decimal places like this:

x = 50/12

x = (INT(x*1000))/1000

print x
print (USING("##.###",x))
print (USING("##.##",x))


Note that this DOES NOT ROUND the value in x, it merely concatenates after the thousandths place to kill the repeating decimal. Keep in mind your new value will be accurate to only TWO decimal places, not three, because the thousandths place was not rounded. The result is 4.166, which eliminates the repeating decimal, and remains accurate to the hundredths (but not the thousandths) place. When you print it with USING, it will round correctly to two places and will display 4.17. If you need greater precision, you can simply concatenate further to the right of the decimal by multiplying and dividing by 10000 or more.

[daved]

Thanks tsh73, I'll keep this code in my templates folder for future use.
Also thanks dan1101.
The code from Walt Decker was exactly what I needed and I used that in my program, which now gives the results I expected.

[Walt Decker]

Although I do not use LB for rounding decimal numbers, here are some functions that can be used:
'
DecNum = 500.765 / 13.176
DecNum = -1 * DecNum
PRINT DecNum
RoundVal = FN.RoundDec(DecNum, 2)
PRINT RoundVal
end

'-----------------------------------
'-----------------------------------

FUNCTION FN.RoundDec(ValIn, NumDecimals)

DecPos  = 0
ValOut  = 0

ValStr$ = ""
Left$   = ""
LftTpl$ = ""
RgtTpl$ = ""
CmbTpl$ = ""

ValStr$ = STR$(ValIn)
DecPos = INSTR(ValStr$, ".")
IF DecPos = 0 THEN
 FN.RoundDec = ValIn
 EXIT FUNCTION
END IF

Left$ = LEFT$(ValStr$, DecPos - 1)
DecPos = LEN(Left$)
LftTpl$ = FN.RepeatStr$(DecPos, "#")
RgtTpl$ = FN.RepeatStr$(NumDecimals, "#")

CmbTpl$ = LftTpl$ + "." + RgtTpl$
ValStr$ = STR$(VAL(USING(CmbTpl$, ValIn)))
ValOut = VAL(ValStr$)

FN.RoundDec = ValOut
END FUNCTION

'-----------------------------------
'-----------------------------------


'-----------------------------------
'-----------------------------------

FUNCTION FN.RepeatStr$(RepNum, RepStr$)

StrOut$ = ""
FOR I = 1 TO RepNum
 StrOut$ = StrOut$ + RepStr$
NEXT I

FN.RepeatStr$ = StrOut$
END FUNCTION
'