Im trying to determine if a power of two is part of a sum (represented by a double). Using a bitwise And works up to a certain point, but then I get an overflow. How can I get this to work with larger numbers?

This works: (2^29 + 2^30) And 2^30

Overflow: (2^29 + 2^31) And 2^30

This works: (2^29 + 2^30) And 2^30

Overflow: (2^29 + 2^31) And 2^30

2^62

2^63 and up are not representable by a Long

Dim D As Double = 2 ^ 65

Dim d2 As Double = D + 2 ^ 29 + D

Debug.Print(d2)

Dim D As Double = 2 ^ 1022

Dim d2 As Double = D + 2 ^ 29 + D

Debug.Print(d2)

so 0001 + 0100 would contain 0010?

i don't think so!

and do that same amount for your bigger number.

and also do that to your compare number

Bitwise Operations can only be performed on integral types so if you can't cast a double value to an integer you can't compare it bitwise.

Dim d3 As Double = 2 ^ 29

Dim d2 As Double = (D + d3)

Debug.Print((D + d3) And 2 ^ 31)

'Returns 2147483648

Dim D As Double = 2 ^ 31

Dim d3 As Double = 2 ^ 29

Dim d2 As Double = (D + d3)

Debug.Print((D + d3) And 2 ^ 29)

'Returns 536870912

Dim D As Double = 2 ^ 31

Dim d3 As Double = 2 ^ 29

Dim d2 As Double = (D + d3)

Debug.Print((D + d3) And 2 ^ 30)

'Returns 0

False = 0 and true = anything non zero

So it works for me Tested in VS 2005

maybey it sounds stupid, but have you tried to convert it to a (binary) string?

7 would be "111"

8 would be "1000"

than you can compare the strings (and you can make it as big as you want)

also you can split the value into more pieces:

yourhugenumber \(16^4) = the right 4 digits removed

yourhugenumber and &HFFFF = only the 4 most right digits

Not the solution you were looking for? Getting a personalized solution is easy.

Ask the Experts2^(x-1) + 2^(x-1)

or when you sum 2^x with any other value

>>

The only time a sum contains 2^x is when you sum

2^(x-1) + 2^(x-1)

or when you sum 2^x with any other value

<<

... unless that "other value" itself contains 2^x, I think ;-)

Corey2

You can get it up to (2^63 + 2^62) AND 2^63 by declaring the variables as UInt64 rather than Double.

PLavelle

I appreciate that the real situation may have been simplified for the purpose of the question but if the situation really is

(2^a + 2^b) AND 2^c

you don't need to be ANDing huge numbers. The "a" sets just one bit. The "b" will not interfere with that setting unless it is equal to "a", in which case it will cancel the original bit and set the next higher bit. If it does not equal to "a" it will set its own bit. The "c" is asking whether a particular bit is set. If that is the bit that was set by "a", and not altered by "b", the answer will be TRUE. If that is the bit that was set by "b" (other than when "b" is equal to "a"), the answer will be TRUE. If "a" did equal "b" and "c" = "a" + 1 (or "b" + 1) the answer will be TRUE. Otherwise, the answer will be FALSE. So you could ignore the "2^" part of the problem and just check the "a", "b" and "c".

As I say, I appreciate that that may be too simplistic for the real problem (and it's really just making PockyMaster's points in another way), but that's my two-pen'orth.

Roger

(2^29 + 2^30) And 2^30

being in the form of

(2^a + 2^b) And 2^c

I can do 2^1023, which gives me a kilobit number. 2^1024 also gives an overflow. This proves that the part of (2^a + 2^b), where (a <= 1023) and (b <= 1022) will never cause an overflow. Nor will the part of 2^c, where c <= 1023

In other words, (2^29 + 2^30) will not cause an overflow, not will 2^30. Also, (2^63 + 2^62) won't cause an overflow.

So, something else must be causing the overflow. All that's left in the expression is the "And" operator. I can't come to any other conclusion that to say that the And operator cannot handle expressions greater than 4 bytes or 32 bits (In Visual Basic that is; it might be 8 bytes or 64 bits in VB.Net).

So I don't believe you can do this by using the And operator. You either need to find another kind of operator, or you need to split the value into separate 32 bits (or 64 bits for .NET) values, as Mark suggested.

I think you're using this kind of binary comparison to check for bit flags, so you don't know the a, b and c values in advance? Otherwise I believe Roger (Sancler) is right by saying that you can use a simple calculation to see wether True or False should be returned.

Cheers,

Luc Derckx

can you do

Dim D As Double

D = 2 ^ 29 + 2 ^ 30

While D > 1

D = D / 2 ^ 30

Wend

If D = 1 Then MsgBox "it is"

That's why I said any OTHER value :D

But I don't know if we are solving a problem here that doesn't exists, simply because binary calculations seem to complicated

Consider the following:

(2^a + 2^b ) AND 2^c

we can apply log 2 to the equation

so, that leaves us with (a + b ) and c

considering that, can't we just do:

public function NumberInSum(ExpA as Integer, ExpB as Integer, ExpC as Integer) as Boolean

'A equals C

if ExpA = ExpC and not ExpB = ExpC then

Return true

'B equals C

elseif ExpB = ExpC and not ExpA = ExpC then

Return true

'ExpA = ExpB and both 1 less than ExpC (causing a shift to the left)

elseif ExpA = (ExpC- 1) ANDALSO ExpB = (ExpC -1) then

return true

else

return false

end if

end function

Now call your function like:

dim bResult as boolean = NumberInSum(29,30,30)

or call like :

NumberInSum ( Log(FirstNumber) / Log(2) , Log(SecondNumber)/Log(2), Log(WhatEverNumberYouWantT

>>

That's why I said any OTHER value :D

<<

4 + 5 ?

5 is an OTHER value than 4. ;-)

That's what comes of being a lawyer for real and only a hobbyist programmer. But, in the specific context of 2^x, I'll grant your point.

How about

If a = b then

return c = a + 1

else

return c = a or c = b

end if

Roger

there are so many ways to write these kind of expressions down, but i tried to keep it simple for PLavelle.

Cheers

>Thanks guys. I've been trying to sort through all this to fully understand what is going on. I'll keep you all updated.

No updates given. I may be a bit harsh, but if PLavelle isn't commenting about those workarounds, nothing but pure logic can say wether those given workarounds are acceptable solutions. I feel that PLavelle already worked it out and forgot about us. I think a couple of us, including me, gave a deeper understanding on why his errors occur in the first place, so why delete this question?

I don't know what others think, but I feel that a points split between Mark_FreeSoftware, Sancler, PockyMaster and myself is appropriate here.

Cheers

Create a free account to continue.

Limited access with a free account allows you to:

- View three pieces of content (articles, solutions, posts, and videos)
- Ask the experts questions (counted toward content limit)
- Customize your dashboard and profile

*This site is protected by reCAPTCHA and the Google Privacy Policy and Terms of Service apply.