Binary Operations

[yumbelie]

Hi,

Could someone explain how using a binary operation, in this case AND is useful:

If I take the binary for 11 and the binary for 13, and AND the results:

1101 (11)
1011 (13)
-----
1001 (9) - Is this TRUE or FALSE?

What exactly does it acheive? I'm asking because a snippet of code I've been given shows binary operations, basically, a number is taken, if the number is negative it is AND'ed with 128 and converted to ASCII Charater equivilent.

When the character string is to be converted back into a number, the string's ASCII Code equivilent is AND'ed with 128, and this is used to set a variable so that we know to make the number negative. I just don't understand how this is working. As I see it (obviously wrongly), if you AND two numbers, unless they are the identical (in binary) the result should be false?

Any help appericated

Thanks

 

[SkipVought]

Hi,

Logical operators perform a bitwise comparison on numbers. AS you can see from your example the 2⁰ and 2³ places are both TRUE -- hence 1001

Skip,

http://www.TheOfficeExperts.com

Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884

 

[SleepDepD]

There are two types of numbers you're dealing with: signed (numbers that can be negative) and unsigned (numbers that are only positive). The range in the ASCII chart is 0-255--this is an 8-bit unsigned number. When you use an 8-bit SIGNED number, the range is seemingly cut in half -128 to 127--still 255 number values though. A signed number is stored with the left-most bit indicating whether the number is positive or negative--when dealing with a 8-bit number, this is the 8th bit, which has a decimal value of 128. In other words, by toggling the 8th bit in a 8-bit signed number, you change it's sign. Hopefully the following illustrates this:

unsigned 8-bit numbers
00000001 = 1
10000001 = 129
11111111 = 255

signed 8-bit numbers
00000001 = 1
10000001 = -1
11111111 = -128

 

[mmilan]

The point here is that the AND is not performed on the actual values, but rather the individual binary bits that go to make that value...

mmilan

 

[yumbelie]

Thanks for the responses, but if say I do: 68 And 63 - does this mean it will return True because the bits required to construct 63 are set? For instance


Chr(68 And 63)

?

Thanks.

 

[Golom]

It will do this
[tt]
68 = 1000100
63 = 0111111
-------
68 AND 63 = 0000100
= 4 (decimal)
[/tt]

 

[yumbelie]

Ah thanks, didn't realise you worked the binary out right to left (etc <-- 8, 4, 2, 1) , not left (1,2,4,8 --> etc) to right.

 

[JeffTullin]

: 68 And 63 - does this mean it will return True because the bits required to construct 63 are set?

If you do anything AND anything, VB will treat any non-zero result as TRUE, if it is inside a bracket.

In some languages, TRUE is 00000001
In some, TRUE is 11111111
This is especially tricky when using (say) VB and C Dlls.
A C DLL may return TRUE as 11111111 which is VB-Speak
for -1
Just something to remember...

 

[SkipVought]

Jeff,

If you do a LOGICAL operation on numbers (A And B, A Or B), the result is a NUMBER, NOT True or False since a bitwise comparison is performed.

Skip,

http://www.TheOfficeExperts.com

Want to get great answers to your Tek-Tips questions? Have a look at FAQ219-2884

 

[strongm]

I rather suspect that Jeff knows this...