Operator MOD

[bplus]

Sorry forgot to get quote for next page, referring to this:
https://qb64phoenix.com/forum/showthread...9#pid10829

Not sure Steve_Mod of the third type will work with Singles, Doubles, _Floats with the Integer Division.

I see precision problems coming back to haunt us with decimal point numbers, the Reals, probably with Steve_Mod of the 2nd type.

[Kernelpanic]

Steve, yours needs some work if you pop a negatives in there as the modulo.

Code: (Select All)

$CONSOLE:ONLY
PRINT "Equation  Steve's Results  |  Pete's Results"
. . .
Function Steve_ModX (num1, num2)
  Steve_ModX = ((num1 Mod num2) + num2) Mod num2
End Function

This gives correct results. What I don't understand is why the program should be wrong because of an addendum? That way I can make any program skid.

Code: (Select All)

Function Steve_ModX (num1, num2)
  Steve_ModX = ((num1 Mod num2) + Abs(num2)) Mod num2
End Function

Steve_ModX = ((num1 Mod num2) + Abs(num2)) Mod num2

[Pete]

Steve updated his function so it would match my routine's capability to work with both positive and negative modulos.

Pete

[Kernelpanic]

Steve updated his function so it would match my routine's capability to work with both positive and negative modulos.

Pete

Oh, and is that correct? 

[SMcNeill]

Steve updated his function so it would match my routine's capability to work with both positive and negative modulos.

Pete

Oh, and is that correct? 

Is 2, or -2, the correct answer to SQR(4)?

BOTH are valid results.  You just need to choose one or the other.  

[vince]

that's both a two and a negative two for all you nobodies

[Kernelpanic]

Oh, and is that correct? 

Is 2, or -2, the correct answer to SQR(4)?

BOTH are valid results.  You just need to choose one or the other.  

Hmm, I do not know! This one: -9 MOD -5 = 1 looks to me: Divides integer with no remainder. Shows the number of b in a.

Code: (Select All)

a = -9: b = -5
c = a \ b
Print c

I wrote a little program to explain this for myself.

Code: (Select All)

'Beispiele fuer MOD - 30. Nov. 2022

Option _Explicit

Dim As Integer a, b, c
Dim As Single f, fganz, frest

a = 13: b = 5

Print

'Liefert den Rest einer ganzzahligen Division
'Returns the remainder of an integer division
c = a Mod b
Print Using "MOD liefer den Rest von 13 MOD 5: ##"; c

Print

'Teilt ganzzahlig ohne Rest. Zeigt die Anzahl von b in a.
'Divides integer with no remainder. Returns the number of b in a.
c = a \ b
Print Using "Anzahl von 5 in 13: ## "; c

Print

'Mit Fliesskommazahlen
f = 7.456
frest = Abs(f - Fix(f))
fganz = f - frest
Print Using "Ganzzahl von 7.456: ##"; fganz
Print Using "Nachkommateil von 7.456: .###"; frest

Print
c = fganz Mod b
Print "Ganzzahl MOD b: ";
Print c

Print
c = Fix(f) Mod b
Print "Und das Ganze direkt mit Fix(f): ";
Print c

End

[Chris]

What problem do You have with implementing a new function to QB64 in the form of:

    MODn(y , x) = y - x * INT(y / x)
    or
     (y MODn x) = y - x * INT(y / x)

All it takes is a little good will. If no one has it, well...
As of today, the MOD is unusable.

Regards - Chris

[SMcNeill]

What problem do You have with implementing a new function to QB64 in the form of:

    MODn(y , x) = y - x * INT(y / x)

What problem do you have in adding and implementing that function into your own code???

As of today, YOUR code is unusable.  Nobody else is having any problem.

[Pete]

>>> MODn(y , x) = y - x * INT(y / x)

Another nice integer modula formula, but I'd tweak it to handle negative mods...

Code: (Select All)

$CONSOLE:ONLY
x = 5
FOR y = 20 TO -20 STEP -1
    PRINT y; MODn(y, x)
NEXT

FUNCTION MODn (y, x)
    MODn = y - ABS(x) * INT(y / ABS(x))
END FUNCTION

Pete