This normalized a bitvec using ofFin
to ofNat
.
msb #
cast #
toInt/ofInt #
zeroExtend and truncate #
Truncating by the bitwidth has no effect.
zero extending a bitvector to width 1 equals the boolean of the lsb.
Zero extending 1#v
to 1#w
equals 1#w
when v > 0
.
Truncating to width 1 produces a bitvector equal to the least significant bit.
extractLsb #
allOnes #
or #
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
and #
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
xor #
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
not #
cast #
shiftLeft #
shiftLeft reductions from BitVec to Nat #
ushiftRight #
ushiftRight reductions from BitVec to Nat #
sshiftRight #
If the msb is true
, the arithmetic shift right equals negating,
then logical shifting right, then negating again.
The double negation preserves the lower bits that have been shifted,
and the outer negation ensures that the high bits are '1'.
sshiftRight reductions from BitVec to Nat #
udiv #
umod #
signExtend #
The sign extension is the same as zero extending when msb = false
.
The sign extension is a bitwise not, followed by a zero extend, followed by another bitwise not
when msb = true
. The double bitwise not ensures that the high bits are '1',
and the lower bits are preserved.
Sign extending to a width smaller than the starting width is a truncation.
Sign extending to the same bitwidth is a no op.
append #
rev #
cons #
Variant of toNat_cons
using +
instead of |||
.
concat #
add #
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
sub/neg #
mul #
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
Equations
- ⋯ = ⋯
le and lt #
ofBoolList #
Rotate Left #
Accessing bits in x.rotateLeft r
the range [0, r)
is equal to
accessing bits x
in the range [w - r, w)
.
Proof by example: Let x := <6 5 4 3 2 1 0> : BitVec 7. x.rotateLeft 2 = (<6 5 | 4 3 2 1 0>).rotateLeft 2 = <3 2 1 0 | 6 5>
(x.rotateLeft 2).getLsbD ⟨i, i < 2⟩ = <3 2 1 0 | 6 5>.getLsbD ⟨i, i < 2⟩ = <6 5>[i] = <6 5 | 4 3 2 1 0>[i + len(<4 3 2 1 0>)] = <6 5 | 4 3 2 1 0>[i + 7 - 2]
Accessing bits in x.rotateLeft r
the range [r, w)
is equal to
accessing bits x
in the range [0, w - r)
.
Proof by example: Let x := <6 5 4 3 2 1 0> : BitVec 7. x.rotateLeft 2 = (<6 5 | 4 3 2 1 0>).rotateLeft 2 = <3 2 1 0 | 6 5>
(x.rotateLeft 2).getLsbD ⟨i, i ≥ 2⟩ = <3 2 1 0 | 6 5>.getLsbD ⟨i, i ≥ 2⟩ = <3 2 1 0>[i - 2] = <6 5 | 3 2 1 0>[i - 2]
Intuitively, grab the full width (7), then move the marker |
by r
to the right (-2)
Then, access the bit at i
from the right (+i)
.
Rotate Right #
Accessing bits in x.rotateRight r
the range [0, w-r)
is equal to
accessing bits x
in the range [r, w)
.
Proof by example: Let x := <6 5 4 3 2 1 0> : BitVec 7. x.rotateRight 2 = (<6 5 4 3 2 | 1 0>).rotateRight 2 = <1 0 | 6 5 4 3 2>
(x.rotateLeft 2).getLsbD ⟨i, i ≤ 7 - 2⟩ = <1 0 | 6 5 4 3 2>.getLsbD ⟨i, i ≤ 7 - 2⟩ = <6 5 4 3 2>.getLsbD i = <6 5 4 3 2 | 1 0>[i + 2]
Accessing bits in x.rotateRight r
the range [w-r, w)
is equal to
accessing bits x
in the range [0, r)
.
Proof by example: Let x := <6 5 4 3 2 1 0> : BitVec 7. x.rotateRight 2 = (<6 5 4 3 2 | 1 0>).rotateRight 2 = <1 0 | 6 5 4 3 2>
(x.rotateLeft 2).getLsbD ⟨i, i ≥ 7 - 2⟩ = <1 0 | 6 5 4 3 2>.getLsbD ⟨i, i ≤ 7 - 2⟩ = <1 0>.getLsbD (i - len(<6 5 4 3 2>) = <6 5 4 3 2 | 1 0> (i - len<6 4 4 3 2>)
When the (i+1)
th bit of x
is false,
keeping the lower (i + 1)
bits of x
equals keeping the lower i
bits.
When the (i+1)
th bit of x
is true,
keeping the lower (i + 1)
bits of x
equalsk eeping the lower i
bits
and then performing bitwise-or with twoPow i = (1 << i)
,
Bitwise and of (x : BitVec w)
with 1#w
equals zero extending x.lsb
to w
.
intMin #
The bitvector of width w
that has the smallest value when interpreted as an integer.
Equations
- BitVec.intMin w = BitVec.twoPow w (w - 1)
Instances For
intMax #
The bitvector of width w
that has the largest value when interpreted as an integer.
Equations
- BitVec.intMax w = BitVec.twoPow w (w - 1) - 1