Trait malachite_base::num::arithmetic::traits::RemPowerOf2

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pub trait RemPowerOf2 {
    type Output;

    // Required method
    fn rem_power_of_2(self, other: u64) -> Self::Output;
}
Expand description

Divides a number by $2^k$, returning just 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 Associated Types§

Required Methods§

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fn rem_power_of_2(self, other: u64) -> Self::Output

Implementations on Foreign Types§

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impl RemPowerOf2 for i8

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fn rem_power_of_2(self, pow: u64) -> i8

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = i8

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impl RemPowerOf2 for i16

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fn rem_power_of_2(self, pow: u64) -> i16

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = i16

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impl RemPowerOf2 for i32

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fn rem_power_of_2(self, pow: u64) -> i32

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = i32

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impl RemPowerOf2 for i64

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fn rem_power_of_2(self, pow: u64) -> i64

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = i64

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impl RemPowerOf2 for i128

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fn rem_power_of_2(self, pow: u64) -> i128

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = i128

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impl RemPowerOf2 for isize

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fn rem_power_of_2(self, pow: u64) -> isize

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

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

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

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = isize

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impl RemPowerOf2 for u8

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fn rem_power_of_2(self, pow: u64) -> u8

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = u8

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impl RemPowerOf2 for u16

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fn rem_power_of_2(self, pow: u64) -> u16

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = u16

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impl RemPowerOf2 for u32

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fn rem_power_of_2(self, pow: u64) -> u32

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = u32

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impl RemPowerOf2 for u64

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fn rem_power_of_2(self, pow: u64) -> u64

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = u64

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impl RemPowerOf2 for u128

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fn rem_power_of_2(self, pow: u64) -> u128

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

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type Output = u128

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impl RemPowerOf2 for usize

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fn rem_power_of_2(self, pow: u64) -> usize

Divides a number by $2^k$, returning just the remainder. For unsigned integers, rem_power_of_2 is equivalent to mod_power_of_2.

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

$$ f(x, k) = x - 2^k\left \lfloor \frac{x}{2^k} \right \rfloor. $$

§Worst-case complexity

Constant time and additional memory.

§Examples

See here.

§

type Output = usize

Implementors§