BitAnd

[Mpro]

Hi! AutoIt developper!

Here's the bug!

The result of this BitAND in AutoIt is -338845696

$Temp = BitAND(3667563225088, 0xFFFFFFFF)

The result is supposed to be = 3956121600

How to explain this result! 1st number too big?

How calculate this number?

Thanks !!!!!!!!

[Richard Robertson]

Yes, it is too big. The first number is a 64 bit integer. BitAnd only works on 32 bit integers.

[Mpro]

Is there a way to overcome this limitation ???

Split my number in half????

[Richard Robertson]

Well it looks like you are trying to get the first 32 bits anyway, so something like this might work.

$number = Dec(Hex(3667563225088))

[martin]

Hi! AutoIt developper!

Here's the bug!

The result of this BitAND in AutoIt is -338845696

$Temp = BitAND(3667563225088, 0xFFFFFFFF)

The result is supposed to be = 3956121600

How to explain this result! 1st number too big?

How calculate this number?

Thanks !!!!!!!!

Wouldn't that be the same as

ConsoleWrite(Mod(3667563225088,2^32 - 1) & @CRLF)

?

But that gives 3956122453 not 3956121600

[martin]

Wouldn't that be the same as

ConsoleWrite(Mod(3667563225088,2^32 - 1) & @CRLF)

?

But that gives 3956122453 not 3956121600

EDIT:second thoughts maybe it should be

ConsoleWrite(Mod(3667563225088,2^32 - 1) - int(3667563225088/(2^32 - 1)) & @CRLF)

[trancexx]

EDIT:second thoughts maybe it should be

ConsoleWrite(Mod(3667563225088,2^32 - 1) - int(3667563225088/(2^32 - 1)) & @CRLF)

Why?

Could it be Mod(3667563225088, 2^32)?

[jchd]

You can do some 64-bit arithmetic using that kludge. You may as well code them with "AutoIt assembler" to get rid of dependency.

[Richard Robertson]

Why?

Could it be Mod(3667563225088, 2^32)?

Mod would be the remainder after division. It wouldn't be the same result as a lower 32 bits mask.

[martin]

Why?

Could it be Mod(3667563225088, 2^32)?

I don't think that is correct because there could be a multiple of 2^32 in the number and that will add to the value from Mod.

Eg

supposing X is less than 2^32 then

Mod(X + 5*(2^32),2^32) = BitAnd(X + 5*(2^32),2^32) + 5

My logic, or lack of it, was like this

BitAnd(16,15) = 0

Mod(16,15) = 1

So I thought that

Mod(A,G) = BitAnd(A,G) + Int(A/G)

EDIT:

Here it is a bit better.

Any number can be written as the sum of 2 numbers like this

D = N + G*K

where N < K and k is 2^p with p any value you like.

BitAnd(D,K-1) is going to be N.

Mod(D,K) = Mod(N + G*K,K)= Mod(Mod(N/K,G) + G),K)=N+G

So

N = Mod(D,K) - G

and G is Int (D/K) Therefore

BitAnd(D,K-1) = Mod(D,K) - Int(D/K)

So now I'm happy that I'm right.

I think I'm happy but I'll check tomorrow.

Edited February 17, 2010 by martin

[Mpro]

I try it in my programm and Mod(3667563225088, 2^32) worked just fine.

Mod(3667563225088, 2^32) = 3956121600

How this one.

BitXOR(3956121600, 1) should be 3956121601

But in Autoit give me -338845695

[jchd]

@Mpro:

If you have repetitive use of such ops, try the small 64-bit code I mentionned above.

@Martin: Hi,

No, your result will be 1 over. Mod(X, Y) returns a result 0 <= Z < Y in all cases. AND 0x00..FF..FF mask won't touch LSB.

[martin]

@Martin: Hi,

No, your result will be 1 over. Mod(X, Y) returns a result 0 <= Z < Y in all cases. AND 0x00..FF..FF mask won't touch LSB.

Hi jchd,

That makes no sense to me! Are you thinking of Div.

I modified my previous post but if you can see an error in my logic please show me what it is.

[jchd]

I don't think that is correct because there could be a multiple of 2^32 in the number and that will add to the value from Mod.

...

Mod(D,K) = Mod(N + G*K,K)= Mod(Mod(N/K,G) + G),K)=N+G

It's there!

By definition of Mod, G*K = 0 mod(K)

So Mod(N + G*K, K) = N , assuming as you did that N < K.

In other words the remainder of the [euclidean] division of kX by X is 0, given X <> 0.

[martin]

It's there!

By definition of Mod, G*K = 0 mod(K)

So Mod(N + G*K, K) = N , assuming as you did that N < K.

In other words the remainder of the [euclidean] division of kX by X is 0, given X <> 0.

I would be surprised if I what I said made sense since my posts last night were hours after I had intended to go to bed.

I think I was thinking that the divisor in Mod was the value to be Anded but it is 1 more obviously. Who knows? Anyway, I was talking bollocks.

Euclidean division? Is that like a Roman division?

[GodlessSinner]

Too big, too small, - size doesn't matter after all:

expandcollapse popup

#include <string.au3>

MsgBox(0, "Custom BitAnd", _BitAnd(3667563225088, 4294967295));  0xFFFFFFFF = 4294967295

;MsgBox(0, "standard", BitAND(134, 172))

Func _BitAnd($arg1,$arg2)
    Local $bin = "", $bin2 = "", $result = ""


While($arg1 >= 1)
    $arg1 /= 2
    If StringIsDigit($arg1) Then
        $bin &= 0
    Else
        $split = StringSplit($arg1, ".")
        $arg1 = $split[1]
        $bin &= 1
    EndIf
WEnd

While($arg2 >= 1)
    $arg2 /= 2
    If StringIsDigit($arg2) Then
        $bin2 &= 0
    Else
        $split = StringSplit($arg2, ".")
        $arg2 = $split[1]
        $bin2 &= 1
    EndIf
WEnd


$bin = _StringReverse($bin)
$bin2 = _StringReverse($bin2)

If StringLen($bin) <> StringLen($bin2) Then
    If StringLen($bin) > StringLen($bin2) Then
        $add =""
        For $i = 1 To (StringLen($bin) - StringLen($bin2)) Step 1
            $add  &= 0
        Next
        $bin2 = $add & $bin2
    Else
        $add = ""
        For $i = 1 To (StringLen($bin2) - StringLen($bin)) Step 1
            $add  &= 0
        Next
        $bin = $add & $bin
    EndIf

EndIf

$split1 = StringSplit($bin,"")
$split2 = StringSplit($bin2,"")
For $i = 1 To StringLen($bin) Step 1
    If $split1[$i] = 0  Or $split2[$i] = 0 Then
        $result &= 0
    ElseIf $split1[$i] = $split2[$i] And $split1[$i] = 1 Then
        $result &= 1
    EndIf
Next


$split = StringSplit($result, "")

$x = $split[1]
For $i = 1 To StringLen($result)-1 Step 1
    $x = $x * 2 + $split[$i+1]
Next
$result = $x
Return $result
EndFunc

Edited February 17, 2010 by Godless

[Mpro]

Thanks Godless for your conversion/AND script.

I will work on XOR script with the same base.

Thanks again!

[jchd]

Euclidean division? Is that like a Roman division?

Just like you learnt it as a kid, remember, long, long ago ;-) and just before they thaught you this shitty decimal point.

A = B * Q + R with A::dividend, B::divisor, Q::quotient, R::remainder and A, Q ∊ ℤ, B ∊ ℤ^*, R ∊ ℕ, R < |Q|

Probably the first equation with two unknowns that one encounters.

[Mpro]

Hi AutoIt developper!

I commented the Godless script to simulate BitAnd & BitXor in Autoit!

This is a great forum with a great peoples!!!

Thanks again Godless!

[Mpro]

Oooooops! Here is the script!

ForumGate.au3