How to calculate the number of digits after the point?

[mestnyi]

For example:

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f.f = 11.1234

On the Internet, write that it is impossible, right?

My attempts

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f.f = 11.1234
Debug StrF(f,4)
Debug StrF(f) 
 
For i=0 To 10
  If Left(StringField(StrF(f), 2, "."),i) <> StringField(StrF(f,i), 2, ".") Or
     Val(Left(Mid(StringField(StrF(f), 2, "."),i+1),1)) = 0
    Break
  EndIf
Next
 
Debug "Count "+i

[Keya]

Im a bit confused by your question, but im wondering if it's the same thing i needed to do only last month which (carefully) trims zero chars off numbers, eg. 1.2300 -> 1.23

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;// Supports Ascii + Unicode

Procedure.s TrimZeros(s$)  ;trim trailing zero chars from a string, eg "1.2300" -> "1.23"
  flen = StringByteLength(s$)
  *pbuf.Ascii = @s$ + flen - SizeOf(Character)
  flen/SizeOf(Character)
  nullpts=flen
  For i = 1 To flen
    If *pbuf\a <> '0' ;And *pbuf\a <> '.' 
      *pbuf + SizeOf(Character)
      *pbuf\a = 0
      Break
    Else
      nullpts-1
    EndIf
    *pbuf - SizeOf(Character)
  Next i
  If nullpts<=0: s$ = "0": EndIf
  If Right(s$,1) = ".": s$ = Left(s$, Len(s$)-1): EndIf
  ProcedureReturn s$
EndProcedure

Procedure xTest(s$)
  Debug s$ + " = " + TrimZeros(s$)
EndProcedure

xTest("40.0")     ;40
xTest("1")        ;1
xTest("12")       ;12
xTest("12.")      ;12
xTest("12.0")     ;12
xTest("0")        ;0
xTest("0.000")    ;0
xTest("0.0")      ;0
xTest("1.123000") ;1.123
xTest("1.1230")   ;1.123
xTest("1.123")    ;1.123
xTest("1.0")      ;1
xTest("1.01")     ;1.01
xTest("1.010")    ;1.01
xTest("1")        ;1
xTest("123")      ;123

[Marc56us]

One way (not the fastest)

String Transform
Find the point then delete it to the left and count
or
Look at the point and then count what remains on the right

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f.f = 11.1234

F$ = StrF(f,4)

j = Len(RemoveString(F$, Left(F$, FindString(F$, "."))))

Debug j ; 4

You can probably do better with pointers

To trim 0, same way

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Procedure xTest(s$)
     If FindString(s$, ".")
          t$ = RTrim(s$, "0")
          t$ = RTrim(t$, ".")
     Else
          t$ = s$
     EndIf
     Debug s$ + " -> " + t$
EndProcedure

xTest("40.0")     ;40
xTest("1")        ;1
xTest("12")       ;12
xTest("12.")      ;12
xTest("12.0")     ;12
xTest("0")        ;0
xTest("0.000")    ;0
xTest("0.0")      ;0
xTest("1.123000") ;1.123
xTest("1.1230")   ;1.123
xTest("1.123")    ;1.123
xTest("1.0")      ;1
xTest("1.01")     ;1.01
xTest("1.010")    ;1.01
xTest("1")        ;1
xTest("123")      ;123 

[mestnyi]

How to calculate the number of digits after the dot?
How can you explain.
I need to know how many digits are in the variable that I specified after the point?

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f.f = 11.1234

;f = ? - Decimals 

StrF(f,Decimals) 

[Demivec]

How to calculate the number of digits after the dot?
How can you explain.
I need to know how many digits are in the variable that I specified after the point?

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f.f = 11.1234

;f = ? - Decimals 

StrF(f,Decimals) 

You can only know the number of decimal digits if the value can be accurately represented by floating point (in binary).

The value in your sample code, 11.1234, is not accurately represented in floating point. So I would say the answer is impossible unless the floating point values are being restricted to only certain values which can be represented in floating point accurately.

[infratec]

What you want is not really possible.

A float variable in real live can have infinite digits after the point.
Example: 1/3 -> 0.3333333333333333333333333333333333333333333333333333333333333333....

How will you count them?

In a computing language a float is limited by the format of 4 used bytes. (14 digits after the point)

If you want the value of the digits which are limited, you have to count the digits until only 0 follows
Use StrF(float, 20) for this, than you are on the safe site

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Define a.f, i.i

For i = 0 To 1000
  Debug Str(i) + " " + StrF(a, 20)
  a + 0.1
Next i

Bernd

[mestnyi]

Impossible is impossible. We will use a string.
I needed to set the increment and get from the increment decimal places

[mestnyi]

What you want is not really possible.

The following code does something like what I needed.

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; Was in the message User_Russian
x.f = 11.1234 

y = Int(x)
z = (x-y)*10000

Debug y
Debug ""+Len(Str(z))+" - "+z

[dcr3]

I dont't know if this is usefull.?

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Dim decimals.s(0)
;D=0
;NAD=NumbersAfterDecimals
NAD.s="11.1234567890123456789012345678901234567890123456789012345678901234567890123456789012"
If FindString(NAD,".")
FNAD = FindString(NAD,".")
decimals(D) = Trim(Right(NAD,Len(NAD)-FNAD))
For Count=0 To Len(decimals(D))-1
Next
            
Debug "NumbersCountAfterDecimal = "+count
EndIf

[Zebuddi123]

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NAD.s="11.1234567890123456789012345678901234567890123456789012345678901234567890123456789012"
x = Len(Mid(NAD, FindString(NAD, Chr(46), 1) + 1))  

Zebuddi.

[dcr3]

Zebuddi. Nice

[Dude]

All the above don't take into account that the decimal separator actually varies by country. A lot of countries use a comma instead, so you need to take that into account by setting up a global variable first:

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NAD.s="11.12345"
x = Len(Mid(NAD, FindString(NAD, Chr(46), 1) + 1))
Debug x ; 8 = Wrong for Denmark, Germany, Greece, Spain, Sweden, etc.

Global decsep$=Space(10)
GetLocaleInfo_(#LOCALE_USER_DEFAULT,#LOCALE_SDECIMAL,decsep$,9)

x = Len(Mid(NAD, FindString(NAD, decsep$, 1) + 1))
Debug x ; 5 = Correct for the above countries. :)

[RASHAD]

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x.f = 11.123456
xstr$ = "11.123456"
nodec = Len(xstr$) - FindString(xstr$, ".")

y = Int(x)
z = (x-y)*Pow(10,nodec)

Debug y
Debug ""+Len(Str(z))+" - "+z

[mestnyi]

What are you doing here?
[

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x.f = 11.123456
xstr$ = "11.123456"
nodec = Len(xstr$) - FindString(xstr$, ".")

y = Int(x)
z = (x-y)*Pow(10,nodec)

Debug y
Debug ""+Len(Str(z))+" - "+z

xstr$ = "11.123456"

this is what from where the string is not known?

Here's what I had in mind: I have 6 digits after the dot sign

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x.f = 11.157498

Procedure.l my(param.f)
  Protected y.i = Int(param)
  Protected z = (param-y)*1000000
  ProcedureReturn z
EndProcedure

z=my(x)
Debug ""+Len(Str(z))+" - "+z