Help with finding largest of 5 random variables

[laughms]

Hey guys, I have a problem with finding the biggest number of 5 (random) variables in these 2 different situations.

Situation 1

$number1= 5

$number2= 7

$number3= 6

$number4= 8

$number5= 2

It does NOT have to tell me that 8 was the biggest, instead I want it to tell me that $number4 was the largest number.

Situation 2

$number1= 7

$number2= 7

$number3= 3

$number4= 7

$number5= 7

In this situation, I just want him to tell me that $number1 was the largest number, and forget about $number 2, 4 and 5. It has to pick out the largest number while choosing in "order". So if $number1 was 5, it had to tell me that $number2 was the biggest and forget about $number4 and $number 5.

Thanks a lot for reading!

 

[Gianni]

Hi laughms

you should use arrays instead of single independant variables (is it allowed? if so)
take a look in the help to this functions:
_ArrayMax
_ArrayMaxIndex
_ArrayUnique

[laughms]

Thanks. I know about _ArrayMax, but if I put those variables as arrays, it will just show me the biggest number. It won't show me which $x was the biggest. Is there a way without arrays?

[BrewManNH]

ArrayMaxIndex will tell you which element holds the highest value.

[jdelaney]

Here ya go:

$number1= 5
$number2= 7
$number3= 6
$number4= 8
$number5= 2

$iCountMax = 0
$iMax = 0
For $i = 1 To 5
    $iNumber = Eval("number" & $i)
    If $iNumber > $iMax Then
        $iMax = $iNumber
        $iCountMax = $i
    EndIf
Next
ConsoleWrite($iMax & @CRLF)
ConsoleWrite("Largest Var = $number" & $iCountMax & "...value=" & Eval("number" & $iCountMax) & @CRLF)

Or, dynamically, for unknown count of vars, or if you ever add more in and don't want to rewrite the code:

$number1= 5
$number2= 7
$number3= 6
$number4= 8
$number5= 2

$iCountMax = 0
$iMax = 0
$i = 1
While True
    $iNumber = Eval("number" & $i)
    If @error Then ExitLoop
    If $iNumber > $iMax Then
        $iMax = $iNumber
        $iCountMax = $i
    EndIf
    $i+=1
Wend
ConsoleWrite($iMax & @CRLF)
ConsoleWrite("Largest Var = $number" & $iCountMax & "...value=" & Eval("number" & $iCountMax) & @CRLF)

both will work for both situations

Edited November 24, 2013 by jdelaney

[Gianni]

 

Here ya go:

$number1= 5
$number2= 7
$number3= 6
$number4= 8
$number5= 2

$iMax = 0
For $i = 1 To 5
    $iNumber = Eval("number" & $i)
    If $iNumber > $iMax Then
        $iMax = $iNumber
        $iCountMax = $i
    EndIf
Next
ConsoleWrite($iMax & @CRLF)
ConsoleWrite("Largest Var = $number" & $iCountMax & "...value=" & Eval("number" & $iCountMax) & @CRLF)

what position was $iMax?

Edit:

I was testing on your first version before you editing (now ok, sorry)

Edited November 24, 2013 by PincoPanco

[jdelaney]

$iMax is the value of the variable...$iCountMax is the integer appended at the end of the variable name.

[Gianni]

$iMax is the value of the variable...$iCountMax is the integer appended at the end of the variable name.

 yes, thanks

just a curiosity, is there a way to have the name of the variable instead of the content? the inverse of eval ()

for example: VarName($iVariable) --> should return "$iVariable" instead of it's content...

[jdelaney]

Not to my knowledge, except for what I have in the second consolewrite.

You would need to assign the variable to a variable, so even if there was such a function, it would return the name of the variable housing it, and not the actual $number#

[Gianni]

Not to my knowledge, except for what I have in the second consolewrite.

You would need to assign the variable to a variable, so even if there was such a function, it would return the name of the variable housing it, and not the actual $number#

 

i mean somethink like this:

Local $iMyVal = 5
ConsoleWrite(VarName($iMyVal) & @CRLF)

Func VarName(ByRef $Var)
    $result = "The referred variable is " & "?? how ??" ; should return "$iMyVal"
    Return $result
EndFunc   ;==>VarName

EDIT:

.... it does not matter, forget it....

excuse me @laughms

sorry for cluttering this thread

Edited November 24, 2013 by PincoPanco

[laughms]

@everybody. Thanks for helping!