Checking to see if @Min is either 1 more or 1 less then Mod(x, 5) = 0

[nitekram]

 

I have no idea why you make that function so complex and redundant while it's actually simple enough to use an almost-oneliner 

- Why are you set on working with an array as a return value while it's possible with an int?

- Why are you duplicating the exact same code in an if-else for cases $modulo >= $n-$range and $modulo <= $range, while the work is the exact same in both cases and you already know that at least one of them is true (because you're inside the first If)?

/edit - Why are you forcing (potentially, i.e. if the minute is not in the accepted ranges) 4 addition and 6 modulo operations while 1 is enough?

- Why are you iniwriting anyway from within a function that has no other work than performing a simple check? That's questionable code flow

 

I guess it all comes down to taste, but my suggestion would be to rethink the whole thing (and then accept my suggestions )

 

At the end of the day it doesn't matter, to each is own If it works for you it works for you. Have a good one!

 

I did most of that code for testing, and to figure out exactly what your code was doing. I wish that I could think your way, as I probably would not be asking these types of questions. It takes me a while to understand abstract concepts, but once I get them, I get them - still not 100% sure on your code, but got this as a step in the right direction?

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n), $iReturn
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        If Mod($value, 5) = 0 Then $iReturn = $value
        If Mod($value + 1, 5) = 0 Then $iReturn = $value + 1
        If Mod($value - 1, 5) = 0 Then $iReturn = $value - 1
        Return $iReturn
    Else
        Return -1
    EndIf
EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

.

[BrewManNH]

That code is far more complex than it needs to be.

[nitekram]

How about this?

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n), $iReturn
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        If $modulo = 0 Then $iReturn = $value
        If $modulo + 1 = 0 Then $iReturn = $value + 1
        If $modulo - 1 = 0 Then $iReturn = $value - 1
        Return $iReturn
    Else
        Return -1
    EndIf
EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

.

EDIT

Or it this better?

Func _isValueWithinRangeFromMultipleOfN($value, $range, $n)
    ; With this function you could do your task by saying _is...(@Min, 1, 5).
    ; But if you decide tomorrow you'd rather check for an interval of 10 minutes with a range of 2 minutes before and after, just do _is...(@Min, 2, 10).
    Local $modulo = Mod($value, $n), $iReturn = -1
    If ($modulo <= $range Or $modulo >= $n - $range) Then
        If $modulo = 0 Then $iReturn = $value
        If $modulo + 1 = 0 Then $iReturn = $value + 1
        If $modulo - 1 = 0 Then $iReturn = $value - 1
    EndIf
    
    Return $iReturn
EndFunc   ;==>_isValueWithinRangeFromMultipleOfN

.

Edited February 19, 2015 by nitekram

[czardas]

Assuming you want to adjust to the nearest 5 minute interval, because your code is hard to interpret and contains bugs, I imagine you are looking for something that does this.

For $i = 0 To 59
    ConsoleWrite(_AdjustToFixedInterval($i, 1, 5, 60) & @LF)
Next

Func _AdjustToFixedInterval($iValue, $iRange, $iInterval, $iMode)
    If $iRange > $iInterval /2 Then Return SetError(1) ; overlap
    Local $modulo = Mod($iValue, $iInterval)
    If $modulo <= $iRange Then Return $iValue - $modulo
    If $modulo >= $iInterval - $iRange Then Return Mod($iValue - $modulo + $iInterval, $iMode)
    Return -1
EndFunc   ;==>_AdjustToFixedInterval

Edited February 19, 2015 by czardas

[nitekram]

 

Assuming you want to adjust to the nearest 5 minute interval, because your code is hard to interpret and contains bugs, I imagine you are looking for something that does this.

For $i = 0 To 59
    ConsoleWrite(_AdjustToFixedInterval($i, 1, 5, 60) & @LF)
Next

Func _AdjustToFixedInterval($iValue, $iRange, $iInterval, $iMode)
    If $iRange > $iInterval /2 Then Return SetError(1) ; overlap
    Local $modulo = Mod($iValue, $iInterval)
    If $modulo <= $iRange Then Return $iValue - $modulo
    If $modulo >= $iInterval - $iRange Then Return Mod($iValue - $modulo + $iInterval, $iMode)
    Return -1
EndFunc   ;==>_AdjustToFixedInterval

I am looking at this code, and though I understand some of it, I am not sure what the last parameters value does to the code - $iMode seems to not do anything, as when I have changed that value to 1, 100, 10, 60 there is no change in the output?

.

[czardas]

Change it to 55 minutes in the hour and you'll see a difference. Also look closer at changing the value to 10. The results are wrong.

;

For $i = 0 To 54 ; 55 minute hour (max input = 54)
    ConsoleWrite(_AdjustToFixedInterval($i, 1, 5, 55) & @LF) ; Altered $iMode to 55
Next

Func _AdjustToFixedInterval($iValue, $iRange, $iInterval, $iMode)
    If $iRange > $iInterval /2 Then Return SetError(1) ; overlap
    Local $modulo = Mod($iValue, $iInterval)
    If $modulo <= $iRange Then Return $iValue - $modulo
    If $modulo >= $iInterval - $iRange Then Return Mod($iValue - $modulo + $iInterval, $iMode)
    Return -1
EndFunc   ;==>_AdjustToFixedInterval

;

Intended as an example, this is very specific to the code you posted, with which you are using @Min. That means the input value will always be below 60. A more universal function would require additional error checks to ensure that the input is not equal to or greater than the mode value, that the mode value is divisible by the interval, that all input values are positive numbers and the mode is greater than 1. I might have overlooked something.

Basically the additional parameter converts the return value 60 to zero, since all the above criteria are met in my >previous example.

Edited February 20, 2015 by czardas

[nitekram]

I see now that there is a difference with those two numbers - and some show up as wrong. Is $iMode in your code the amount of minutes, if not what would be the definition of that variable?

[czardas]

Yes it's the number of minutes in an hour. The adjustment of the result from 59 to 60 should probably return a valid (zero) number of minutes past the hour, and 60 is not a valid number of minutes past the hour. If you wanted to return 60 and deal with the overflow outside the function, it is an easy modification to make as shown below.

With only three parameters the last return value becomes 60, which really should be zero because it's the start of the next hour. The best solution will depend on exactly what you are intending to do.

For $i = 0 To 59
    ConsoleWrite(_AdjustToFixedInterval($i, 1, 5) & @LF)
Next

Func _AdjustToFixedInterval($iValue, $iRange, $iInterval)
    If $iRange > $iInterval /2 Then Return SetError(1) ; overlap
    Local $modulo = Mod($iValue, $iInterval)
    If $modulo <= $iRange Then Return $iValue - $modulo
    If $modulo >= $iInterval - $iRange Then Return $iValue - $modulo + $iInterval
    Return -1
EndFunc   ;==>_AdjustToFixedInterval

Edited February 20, 2015 by czardas

[nitekram]

Thanks for everyone's help.