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 Author Topic: HELP: Why does this not work?  (Read 9119 times)
Moneo
Na_th_an

Posts: 1971

 « on: July 14, 2003, 01:40:49 PM »

The following anomaly occurs in QuickBasic 4.5. Does anyone know why this happens?
Code:

DEFLNG a-z
dim a as integer
dim n as integer

n = 128

rem The following two instructions work perfectly:

a=log(n)/log(2)
if 2^a = n then print n, "is a power of 2"

rem The following instruction, which is a combination of the above, DOES NOT WORK.

if 2^(log(n)/log(2)) = n then print n, "is a power of 2"

I don't understand. Why does it work separately but not combined?
Thanks.
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na_th_an
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Posts: 8244

 « Reply #1 on: July 14, 2003, 01:44:15 PM »

Dunno, but maybe the result of log(n)/log(2) isn't integer, hence the error.

a is integer, so a = log(n)/log(2) will perform an implicit INT(log(n)/log(2)), and, generally:

Code:
2^INT(log(n)/log(a))    <>    2^(log(n)/log(a))

Check it out
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Moneo
Na_th_an

Posts: 1971

 « Reply #2 on: July 14, 2003, 02:23:06 PM »

Nathan,
So if I changed the failing instruction as follows, it should work.

if 2^INT(log(n)/log(2)) = n then print n, "is a power of 2"

WELL, IT DOES NOT WORK. ?
Any ideas?
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na_th_an
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 « Reply #3 on: July 14, 2003, 02:24:25 PM »

Well, this time I HAVE NO CLUE!!  ::  Another QB bug?
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Glenn
I hold this place together

Posts: 786

 « Reply #4 on: July 14, 2003, 02:49:56 PM »

It fails at n = 16384.  However, changing INT to CINT makes it work for 16384.  (CINT converts to INTEGER and rounds correctly.  INT just truncates.)
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ravelling Curmudgeon
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Moneo
Na_th_an

Posts: 1971

 « Reply #5 on: July 14, 2003, 02:55:26 PM »

Glenn,
If I put into a loop, it also works up to 1024,
EXCEPT THAT 128 DOESN'T WORK.

When n=128, the one line combined instruction does not work with or without the INT.
If I round the result inside the parentesis, it will work. But why do I need to do this when all the other numbers work without it?
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Glenn
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Posts: 786

 « Reply #6 on: July 14, 2003, 03:04:51 PM »

don't store floating point numbers accurately (and LOG returns a floating point result).  When n = 128, LOG(n) / log(2) should be 7 but what you get is something slightly less than 7.  Similarly, if n = 16384, the result should be 14 but you get something slightly less than 14.  So, INT truncates the result to 6 or 13.   For the other values, you get a result slightly above the integer that math says you should get and so INT's truncation works.  Instead of using CINT, the perhaps more insightful solution is to use

INT(LOG(n) / LOG(2) + .001)
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ravelling Curmudgeon
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Agamemnus
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Posts: 3491

 « Reply #7 on: July 14, 2003, 03:38:45 PM »

or just use code to get the max bits used.
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Glenn
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Posts: 786

 « Reply #8 on: July 14, 2003, 03:45:06 PM »

That right value isn't known a priori.  He's in fact using code to get the number of bits used.
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ravelling Curmudgeon
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Windows should be defenestrated.
Moneo
Na_th_an

Posts: 1971

 « Reply #9 on: July 14, 2003, 03:46:13 PM »

Quote from: "Agamemnus"
or just use code to get the max bits used.

Sorry, I don't know what you mean.
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Agamemnus
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Posts: 3491

 « Reply #10 on: July 14, 2003, 03:59:43 PM »

128 uses 7 bits. 127 uses 7 bits. If 127 is less than 128, 127 isn't a power of 2.
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Moneo
Na_th_an

Posts: 1971

 « Reply #11 on: July 14, 2003, 04:08:45 PM »

Quote from: "Agamemnus"
128 uses 7 bits. 127 uses 7 bits. If 127 is less than 128, 127 isn't a power of 2.

I agree with your analysis, but how do I apply it to solving my posted problem?
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Glenn
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Posts: 786

 « Reply #12 on: July 14, 2003, 04:11:50 PM »

truly amazing.
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ravelling Curmudgeon
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I liked spam better when it was something that came in a can.
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Agamemnus
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Posts: 3491

 « Reply #13 on: July 14, 2003, 04:20:45 PM »

because 2 4 8 16 32 64 128 256 512 1024 2048 4096 8192 and 16384 are all powers of two, and code (that I forgot) that gets the max bits used for a number (how many bits are needed to represent it) that is then subtracted from 2^those bits (or the array in the beginning) if equals zero, then it is a power of two.

Or, do this:

Code:

sub power.of.two% (n%):
select case n%
case 2, 4 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384
power.of.two% = -1
end select
end function
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Moneo
Na_th_an

Posts: 1971

 « Reply #14 on: July 14, 2003, 06:19:56 PM »

AGAMEMNUS,
I know you're tying to help, but I'm not looking for alternate ways of determining if a number is a power of 2. I can already do that in the following 2 lines:
Code:

DEFLNG a-z
dim a as integer
dim n as integer

a=log(n)/log(2)
if 2^a = n then print n, "is a power of 2"

My problem is that I wanted to do it one line of code like this:
Code:
if 2^(log(n)/log(2)) = n then print n, "is a power of 2"

But, for some reason it doesn't work when n=128.
The purpose of my post was to ask why this doesn't work. Nathan had a good idea, but it didn't solve the problem.
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