Improve PartialEq for slices

Exploiting the fact that getting the length of the slices is known, we
can use a counted loop instead of iterators, which means that we only
need a single counter, instead of having to increment and check one
pointer for each iterator.

Benchmarks comparing vectors with 100,000 elements:

Before:

```
running 8 tests
test eq1_u8  ... bench:      66,757 ns/iter (+/- 113)
test eq2_u16 ... bench:     111,267 ns/iter (+/- 149)
test eq3_u32 ... bench:     126,282 ns/iter (+/- 111)
test eq4_u64 ... bench:     126,418 ns/iter (+/- 155)
test ne1_u8  ... bench:      88,990 ns/iter (+/- 161)
test ne2_u16 ... bench:      89,126 ns/iter (+/- 265)
test ne3_u32 ... bench:      96,901 ns/iter (+/- 92)
test ne4_u64 ... bench:      96,750 ns/iter (+/- 137)
```

After:

```
running 8 tests
test eq1_u8  ... bench:      46,413 ns/iter (+/- 521)
test eq2_u16 ... bench:      46,500 ns/iter (+/- 74)
test eq3_u32 ... bench:      50,059 ns/iter (+/- 92)
test eq4_u64 ... bench:      54,001 ns/iter (+/- 92)
test ne1_u8  ... bench:      47,595 ns/iter (+/- 53)
test ne2_u16 ... bench:      47,521 ns/iter (+/- 59)
test ne3_u32 ... bench:      44,889 ns/iter (+/- 74)
test ne4_u64 ... bench:      47,775 ns/iter (+/- 68)
```
This commit is contained in:
Björn Steinbrink 2015-07-08 14:49:55 +02:00
parent 7b7853897b
commit 9f4d5b4be1

View File

@ -1459,12 +1459,30 @@ pub mod bytes {
#[stable(feature = "rust1", since = "1.0.0")] #[stable(feature = "rust1", since = "1.0.0")]
impl<A, B> PartialEq<[B]> for [A] where A: PartialEq<B> { impl<A, B> PartialEq<[B]> for [A] where A: PartialEq<B> {
fn eq(&self, other: &[B]) -> bool { fn eq(&self, other: &[B]) -> bool {
self.len() == other.len() && if self.len() != other.len() {
order::eq(self.iter(), other.iter()) return false;
}
for i in 0..self.len() {
if !self[i].eq(&other[i]) {
return false;
}
}
true
} }
fn ne(&self, other: &[B]) -> bool { fn ne(&self, other: &[B]) -> bool {
self.len() != other.len() || if self.len() != other.len() {
order::ne(self.iter(), other.iter()) return true;
}
for i in 0..self.len() {
if self[i].ne(&other[i]) {
return true;
}
}
false
} }
} }