Real --> integer_part + decimal_part

[Richard]

Looking for best way to do ...

I will process the integer_part and decimal_part differently




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[jack]

Richard, I would try Int to get the integer part, then for the fractional part simply subtract the integer part
you need to be aware that Int behaves like C floor so for example

Code: QB64: [Select]

1. Dim As _Float y
2. Dim As _Integer64 k
3.  
4. y = -3141592653589.7932384626433832795##
5. k = Int(y)
6. Print k ' -3141592653590
7.  

you could also use C runtime functions like trunc, trunc will simply truncate the number so if you used trunc in the above example you would get -3141592653589
what range and precision are the values you want to work with?

[jack]

for example

Code: QB64: [Select]

1. Option _Explicit
2. Declare Library
3.     Function trunc# (ByVal x As Double)
4. End Declare
5.  
6. Dim As Double y
7. Dim As _Integer64 k
8.  
9. y = -3141592653589.7932384626433832795#
10. k = trunc(y)
11. Print y
12. Print k, y - k
13.  

[SMcNeill]

Why do you need a declare library for that, jack?

k = INT(y)

Why doesn’t the above work?  Am I missing something here?

[jack]

@SMcNeill you apparently didn't read my post carefully, Int(-3.5) gives -4, Trunc(-3.5) gives -3
you could of course work around Int's behavior maybe sgn(x)*int(abs(x))

[FellippeHeitor]

Try Fix()

[jack]

right, Fix will work

[Richard]

I still have trouble with the decimal_part.

@jack the "displayed precision" may only require LONG (INTEGER may be a bit too small) and anticipated only 3 decimal values (at most 4) but the REAL NUMBER may be FLOAT




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[jack]

@jack ... but the REAL NUMBER may be FLOAT

in that case don't use _Float, if you use Double then you should get the expected result, as it seems there's a bug in Print Using with _Float
edit, sorry I took FLOAT as meaning Single, for _Float you would need to work around the bug

[SMcNeill]

Have you tried simply making it a string and just separating it at the decimal?

s$ = STR(x#)
decimal = INSTR(s$, “.”
leftside = LEFT$(s$, decimal)
rightside = MID$(s$, decimal +1)

[jack]

something anyone that uses the type _Float should be aware of is that only basic arithmetic operators are supported, that is +, -, *,  / and <, =, >, <>
not even Abs() is supported the functions return double precision result, except for Exp()

[jack]

@Richard
may I suggest something like this

Code: Text: [Select]

1. Option _Explicit
2. Declare Library
3.     Function snprintf& (Dest As String, Byval l As Long, frmt As String, Byval x As _Float)
4. End Declare
5.  
6. Dim As _Float c
7.  
8. c = 123.45644444444444##
9. Print strf(c)
10. Function strf$ (x As _Float)
11.     Dim As String s
12.     Dim As String frmt
13.     Dim As Long l, sign
14.     sign = Sgn(x)
15.     If sign < 0 Then x = -x
16.     s = Spc(64)
17.     frmt = "%19.3Lf" + Chr$(0)
18.     l = snprintf(s, Len(s), frmt, x)
19.     s = _Trim$(s)
20.     If sign < 0 Then s = "-" + s Else s = " " + s
21.     strf = s
22. End Function
23.  

[Richard]

@FellippeHeitor @SMcNeill @jack  Many thanks



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@jack I will try out your last reply soon - it was only the displaying on screen what I was trying to sort out. Thanks for the IMPORTANT REMINDER re where _FLOAT precision is maintained.

[jack]

Richard, if you need special functions evaluated to _Float precision you can call the C runtime functions https://en.cppreference.com/w/c/numeric/math
for example

Code: QB64: [Select]

1. Declare Library
2.     Function snprintf& (Dest As String, Byval l As Long, frmt As String, Byval x As _Float)
3.     Function logl## (ByVal x As _Float)
4.     Function expl## (ByVal x As _Float)
5.     Function sinl## (ByVal x As _Float)
6.     Function cosl## (ByVal x As _Float)
7.     Function tanl## (ByVal x As _Float)
8.     Function asinl## (ByVal x As _Float)
9.     Function acosl## (ByVal x As _Float)
10.     Function atanl## (ByVal x As _Float)
11.     Function tgammal## (ByVal x As _Float)
12.     Function truncl## (ByVal x As _Float)
13.     Function fabsl## (ByVal x As _Float)
14. End Declare
15.