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