Auto merge of #26884 - dotdash:fast, r=alexcrichton

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)
```

Author: bors

src/libcore/slice.rs

@@ -1463,12 +1463,30 @@ pub mod bytes {
 #[stable(feature = "rust1", since = "1.0.0")]
 impl<A, B> PartialEq<[B]> for [A] where A: PartialEq<B> {
     fn eq(&self, other: &[B]) -> bool {
-        self.len() == other.len() &&
-            order::eq(self.iter(), other.iter())
+        if self.len() != other.len() {
+            return false;
+        }
+
+        for i in 0..self.len() {
+            if !self[i].eq(&other[i]) {
+                return false;
+            }
+        }
+
+        true
     }
     fn ne(&self, other: &[B]) -> bool {
-        self.len() != other.len() ||
-            order::ne(self.iter(), other.iter())
+        if self.len() != other.len() {
+            return true;
+        }
+
+        for i in 0..self.len() {
+            if self[i].ne(&other[i]) {
+                return true;
+            }
+        }
+
+        false
     }
 }