Fixed bug when initialising bitv from init=true #11615
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| @@ -269,14 +269,23 @@ impl Bitv { | |||
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| impl Bitv { | |||
| pub fn new(nbits: uint, init: bool) -> Bitv { | |||
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Can you add documentation for new (ie, what the parameters do)
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Nice find! |
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Jan 19, 2014
This is my first patch so feedback appreciated! Bug when initialising `bitv:Bitv::new(int,bool)` when `bool=true`. It created a `Bitv` with underlying representation `!0u` rather than the actual desired bit layout ( e.g. `11111111` instead of `00001111`). This works OK because a size attribute is included which keeps access to legal bounds. However when using `BitvSet::from_bitv(Bitv)`, we then find that `bitvset.contains(i)` can return true when `i` should not in fact be in the set. ``` let bs = BitvSet::from_bitv(Bitv::new(100, true)); assert!(!bs.contains(&127)) //fails ``` The fix is to create the correct representation by treating various cases separately and using a bitshift `(1<<nbits) - 1` to generate correct number of `1`s where necessary.
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This is my first patch so feedback appreciated!
Bug when initialising
bitv:Bitv::new(int,bool)whenbool=true. It created aBitvwith underlying representation!0urather than the actual desired bit layout ( e.g.11111111instead of00001111). This works OK because a size attribute is included which keeps access to legal bounds. However when usingBitvSet::from_bitv(Bitv), we then find thatbitvset.contains(i)can return true whenishould not in fact be in the set.The fix is to create the correct representation by treating various cases separately and using a bitshift
(1<<nbits) - 1to generate correct number of1s where necessary.