why does Labview take 8 bits to represent a Boolean?

[AJ_CS]

Since it can take only one of two values, can it not be represented by a bit?

[Yamaeda · Accepted solution]

The memory controllers smallest unit is a byte, and reading from memory is done 32 bit at a time, so there's no benefit to use a single bit. If you use alot of bits, you can mask them into an I32 which is often done in e.g. C.

/Y

[AJ_CS]

OK..I get it..I also felt there would be some thing like smallest unit or so..thank for clarifying..

[Yamaeda]

The benefit of booleans is that even when you're wrong, you're only a bit off. 😉

/Y

[crossrulz]

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@Yamaeda wrote:

The benefit of booleans is that even when you're wrong, you're only a bit off. 😉

/Y

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Unless you are having fun with Type Casting

0 = False

Everything Else = True

[Yamaeda]

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@crossrulz wrote:

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@Yamaeda wrote:

The benefit of booleans is that even when you're wrong, you're only a bit off. 😉

/Y

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Unless you are having fun with Type Casting

0 = False

Everything Else = True

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I guess that's like arguing with girlfriend/wife; when all bits align you're right, else you're wrong! 😄

/Y

[johnsold]

Early versions of LV did use packed bits but as processors became more powerful and memory became cheaper the overhead to pack and upack the bits outweighed the smaller memory footprint.

Lynn

[rolfk]

Only for boolean arrays. And skalar booleans were 16 bit which was the standard boolean format on MacOS 6 + 7. The complication and performance loss with packing and unpacking, made the LabVIEW developers change the boolean representation in LabVIEW 5 to be byte sized, like most C++ compilers use too, and to forget about packing for boolean arrays.

[johnsold]

Rolf,

Thank you for correcting the details.

Lynn