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feat: use the bit manipulation to compute the alignment size #1206

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Aug 5, 2024
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61 changes: 44 additions & 17 deletions commons/zenoh-shm/src/api/provider/types.rs
Original file line number Diff line number Diff line change
Expand Up @@ -52,7 +52,7 @@ impl Display for AllocAlignment {
impl Default for AllocAlignment {
fn default() -> Self {
Self {
pow: (std::mem::align_of::<u32>() as f64).log2().round() as u8,
pow: std::mem::align_of::<u32>().ilog2() as _,
}
}
}
Expand All @@ -65,9 +65,10 @@ impl AllocAlignment {
/// This function will return an error if provided alignment power cannot fit into usize.
#[zenoh_macros::unstable_doc]
pub const fn new(pow: u8) -> Result<Self, ZLayoutError> {
match pow {
pow if pow < usize::BITS as u8 => Ok(Self { pow }),
_ => Err(ZLayoutError::IncorrectLayoutArgs),
if pow < usize::BITS as u8 {
Ok(Self { pow })
} else {
Err(ZLayoutError::IncorrectLayoutArgs)
}
}

Expand All @@ -92,19 +93,45 @@ impl AllocAlignment {
/// ```
#[zenoh_macros::unstable_doc]
pub fn align_size(&self, size: NonZeroUsize) -> NonZeroUsize {
let alignment = self.get_alignment_value();
match size.get() % alignment {
0 => size,
// SAFETY:
// This unsafe block is always safe:
// 1. 0 < remainder < alignment
// 2. because of 1, the value of (alignment.get() - remainder) is always > 0
// 3. because of 2, we add nonzero size to nonzero (alignment.get() - remainder) and it is always positive if no overflow
// 4. we make sure that there is no overflow condition in 3 by means of alignment limitation in `new` by limiting pow value
remainder => unsafe {
NonZeroUsize::new_unchecked(size.get() + (alignment.get() - remainder))
},
}
// Notations:
// - size to align S
// - usize::BITS B
// - pow P where 0 ≤ P < B
// - alignment value A = 2^P
// - return R = min{x | x ≥ S, x % A = 0}
//
// Example 1: A = 4 = (00100)₂, S = 4 = (00100)₂ ⇒ R = 4 = (00100)₂
// Example 2: A = 4 = (00100)₂, S = 7 = (00111)₂ ⇒ R = 8 = (01000)₂
// Example 3: A = 4 = (00100)₂, S = 8 = (01000)₂ ⇒ R = 8 = (01000)₂
// Example 4: A = 4 = (00100)₂, S = 9 = (01001)₂ ⇒ R = 12 = (01100)₂
//
// Algorithm: For any x = (bₙ, ⋯, b₂, b₁)₂ in binary representation,
// 1. x % A = 0 ⇔ ∀i < P, bᵢ = 0
// 2. f(x) ≜ x & !(A-1) leads to ∀i < P, bᵢ = 0, hence f(x) % A = 0
// (i.e. f zeros all bits before the P-th bit)
// 3. R = min{x | x ≥ S, x % A = 0} is equivalent to find the unique R where S ≤ R < S+A and R % A = 0
// 4. x-A < f(x) ≤ x ⇒ S-1 < f(S+A-1) ≤ S+A-1 ⇒ S ≤ f(S+A-1) < S+A
//
// Hence R = f(S+A-1) = (S+(A-1)) & !(A-1) is the desired value

// Compute A - 1 = 2^P - 1
let a_minus_1 = self.get_alignment_value().get() - 1;

// Overflow check: ensure S ≤ 2^B - 2^P = (2^B - 1) - (A - 1)
// so that R < S+A ≤ 2^B and hence it's a valid usize
let bound = usize::MAX - a_minus_1;
assert!(
size.get() <= bound,
"The given size {} exceeded the maximum {}",
size.get(),
bound
);

// Overflow never occurs due to the check above
let r = (size.get() + a_minus_1) & !a_minus_1;

// SAFETY: R ≥ 0 since R ≥ S ≥ 0
unsafe { NonZeroUsize::new_unchecked(r) }
}
}

Expand Down