Trait malachite_base::num::arithmetic::traits::ModPowerOf2Assign
source · [−]pub trait ModPowerOf2Assign {
fn mod_power_of_2_assign(&mut self, other: u64);
}
Expand description
Divides a number by $2^k$, replacing the number by the remainder. The remainder is non-negative.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
Required Methods
fn mod_power_of_2_assign(&mut self, other: u64)
Implementations on Foreign Types
sourceimpl ModPowerOf2Assign for u8
impl ModPowerOf2Assign for u8
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for u16
impl ModPowerOf2Assign for u16
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for u32
impl ModPowerOf2Assign for u32
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for u64
impl ModPowerOf2Assign for u64
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for u128
impl ModPowerOf2Assign for u128
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for usize
impl ModPowerOf2Assign for usize
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder.
If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Examples
See here.
sourceimpl ModPowerOf2Assign for i8
impl ModPowerOf2Assign for i8
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.
sourceimpl ModPowerOf2Assign for i16
impl ModPowerOf2Assign for i16
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.
sourceimpl ModPowerOf2Assign for i32
impl ModPowerOf2Assign for i32
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.
sourceimpl ModPowerOf2Assign for i64
impl ModPowerOf2Assign for i64
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.
sourceimpl ModPowerOf2Assign for i128
impl ModPowerOf2Assign for i128
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.
sourceimpl ModPowerOf2Assign for isize
impl ModPowerOf2Assign for isize
sourcefn mod_power_of_2_assign(&mut self, pow: u64)
fn mod_power_of_2_assign(&mut self, pow: u64)
Divides a number by $2^k$, replacing the first number by the remainder. The remainder is non-negative.
If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.
$$ x \gets x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$
Worst-case complexity
Constant time and additional memory.
Panics
Panics if self
is negative and pow
is greater than or equal to Self::WIDTH
.
Examples
See here.