Trait malachite_base::num::arithmetic::traits::RemPowerOf2Assign

pub trait RemPowerOf2Assign {
    // Required method
    fn rem_power_of_2_assign(&mut self, other: u64);
}

Divides a number by $2^k$, replacing the number by the remainder. The remainder has the same sign as the number.

If the quotient were computed, the quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq |r| < 2^k$.

Required Methods

fn rem_power_of_2_assign(&mut self, other: u64)

Implementations on Foreign Types

impl RemPowerOf2Assign for i8

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for i16

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for i32

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for i64

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for i128

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for isize

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. The remainder has the same sign as the first number.

If the quotient were computed, he quotient and remainder would satisfy $x = q2^k + r$ and $0 \leq r < 2^k$.

$$ x \gets x - 2^k\operatorname{sgn}(x)\left \lfloor \frac{|x|}{2^k} \right \rfloor. $$

Worst-case complexity

Constant time and additional memory.

Examples

See here.

impl RemPowerOf2Assign for u8

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

impl RemPowerOf2Assign for u16

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

impl RemPowerOf2Assign for u32

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

impl RemPowerOf2Assign for u64

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

impl RemPowerOf2Assign for u128

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

impl RemPowerOf2Assign for usize

fn rem_power_of_2_assign(&mut self, pow: u64)

Divides a number by $2^k$, replacing the first number by the remainder. For unsigned integers, rem_power_of_2_assign is equivalent to mod_power_of_2_assign.

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.

Implementors