Converting string to float

[marcoagpinto]

Hello!

Code: Select all

t$="0.60"
Debug ValF(t$)

Instead of 0.60 it says 0.60000002384186.

How do I get 0.60?

Thanks!

[infratec]

Code: Select all

t$="0.60"
Debug StrF(ValF(t$), 2)

[TI-994A]

How do I get 0.60?!

Code: Select all

t$ = "0.60"
t.f = ValF(t$)
Debug StrF(t, 2)

[tastyraspberry]

It doesn't convert exactly because some decimal numbers can't be represented in binary exactly eg 0.1 expands to 0.000110011001100110011 recurring..

http://www.exploringbinary.com/why-0-po ... ing-point/

I presume as 0.6 is 0.1 * 6 it is why you get the 0.60000002384186. it is the closest that can be done in binary.

It has big implications for financial software and you should always represent money as integers of the smallest division ie pence, cent etc and then just divide by 100 to represent the pound/euro/dollar.

Michael

[IdeasVacuum]

Also, you can use Double instead of Float (unless using an API function that requires float)

[captain_skank]

It has big implications for financial software and you should always represent money as integers of the smallest division ie pence, cent etc and then just divide by 100 to represent the pound/euro/dollar.

Michael

This is true, but I've had trouble in the past displaying the result becasue you have to convert back to a string to do so ( using StrF or StrD ) and would love for PB to have currency handling functions built in.

[citystate]

It has big implications for financial software and you should always represent money as integers of the smallest division ie pence, cent etc and then just divide by 100 to represent the pound/euro/dollar.

Michael

This is true, but I've had trouble in the past displaying the result becasue you have to convert back to a string to do so ( using StrF or StrD ) and would love for PB to have currency handling functions built in.

for decimal currencies (ie Pounds, Euros, Dollars), perhaps something like this to displaying the integer?

Code: Select all

currency.l = 12345 ; $123.45
output$ = Str(currency/100) + "." + Str(currency%100)
debug output$

[marcoagpinto]

Thanks, guys!

[davido]

@citystate,

May I suggest the slight modification to take care of less than 10 cents.

Code: Select all

currency.l = 12305 ; $123.05
output$ = Str(currency/100) + "." + RSet(Str(currency%100),2,"0")
Debug output$[/quote]