Can someone explain?

[Dave Young]

I'm looking at some code that i've inherited and I'm not really familar with
what's going on here and was hoping somone could explain it to me.

For reference:

f1 is a long
f2 is a long

here's the statement

f1 = (f1 or f2)

I'm curious what the "OR" is doing here. I assume it's some type of
comparison and reassignment based on the comparison but like I said, I'm not
sure I've ever seen this type of operation before and would like someone to
try and explain what's going on here.

Thanks

[Armin Zingler]

"Dave Young" <dm******@yahoo.comschrieb

f1 = (f1 or f2)

I'm curious what the "OR" is doing here. I assume it's some type of

There's the helpful [F1] key on your keyboard. Can lead you to:

http://msdn.microsoft.com/en-us/library/06s37a7f.aspx
Armin

[Rory Becker]

Hello Dave,

For reference:

f1 is a long
f2 is a long
here's the statement

f1 = (f1 or f2)

I'm curious what the "OR" is doing here. I assume it's some type of
comparison and reassignment based on the comparison but like I said,
I'm not sure I've ever seen this type of operation before and would
like someone to try and explain what's going on here.

Imagine that before this statement the following is true.
f1 = 7 = 00000111 (bin 7)
f2 = 24 = 00011000 (bin 24)

the statement says set f1 equal to a value equivilent to the ORing of both
values.

ie if the f1 has a 1 in a given position OR f2 has a 1 in that same position
then the output value should have a 1 in that same position

f1 therefore becomes 00011111 = 31

in this case it appears to have resulted in addition

However.... imagine that the figures were 15 and 28
15 = 00001111
28 = 00011100
f1 = 00011111 = 32

32 is definitly not the result of adding 15 and 28

It can be looked on as the combination of flags.

It looks like in you example code.... someone was trying to flip all the
bit on in f1 which were on in f2 without turning off any of the bits which
were already on in f1.

Also be on the lookout for code which uses the AND (bitwise) operator

This sets the output bits only where both input bits are 1. this is used
for finding out the intersection of 2 sets of flags (ie what they have in
common)

It's late here. I hope some of this makes sense. :)

--
Rory

[Dave Young]

Thanks Rory

Guess it's time to study up on bitwise operators.

"Rory Becker" <ro********@newsgroup.nospamwrote in message
news:3a**************************@news.microsoft.c om...

Hello Dave,

>For reference:

f1 is a long
f2 is a long
here's the statement

f1 = (f1 or f2)

I'm curious what the "OR" is doing here. I assume it's some type of
comparison and reassignment based on the comparison but like I said,
I'm not sure I've ever seen this type of operation before and would
like someone to try and explain what's going on here.

Imagine that before this statement the following is true.
f1 = 7 = 00000111 (bin 7)
f2 = 24 = 00011000 (bin 24)

the statement says set f1 equal to a value equivilent to the ORing of both
values.

ie if the f1 has a 1 in a given position OR f2 has a 1 in that same
position then the output value should have a 1 in that same position

f1 therefore becomes 00011111 = 31

in this case it appears to have resulted in addition

However.... imagine that the figures were 15 and 28
15 = 00001111
28 = 00011100
f1 = 00011111 = 32

32 is definitly not the result of adding 15 and 28

It can be looked on as the combination of flags.

It looks like in you example code.... someone was trying to flip all the
bit on in f1 which were on in f2 without turning off any of the bits which
were already on in f1.

Also be on the lookout for code which uses the AND (bitwise) operator

This sets the output bits only where both input bits are 1. this is used
for finding out the intersection of 2 sets of flags (ie what they have in
common)

It's late here. I hope some of this makes sense. :)

--
Rory

[Cor Ligthert[MVP]]

Dave,

Guess it's time to study up on bitwise operators.

Keep in mind that they are from the age that memory (in fact bits) were very
expensive.

Cor

[Mythran]

"Cor Ligthert[MVP]" <no************@planet.nlwrote in message
news:B2**********************************@microsof t.com...

Dave,

>Guess it's time to study up on bitwise operators.

Keep in mind that they are from the age that memory (in fact bits) were
very expensive.

Cor

That being said, you can come to the same results using the FlagsAttribute
on an enumeration.

Mythran

[Chris Dunaway]

On Jul 1, 4:39 pm, Rory Becker <[e-mail]:

However.... imagine that the figures were 15 and 28
15 = 00001111
28 = 00011100
f1 = 00011111 = 32

32 is definitly not the result of adding 15 and 28
It's late here. I hope some of this makes sense. :)

Must be why you got 32 out of that OR rather than 31 which is the
correct answer! :)

Chris