pub fn memrchr2(needle1: u8, needle2: u8, haystack: &[u8]) -> Option<usize>Expand description
Search for the last occurrence of two possible bytes in a haystack.
This returns the index corresponding to the last occurrence of one of the
needle bytes in haystack, or None if one is not found. If an index is
returned, it is guaranteed to be less than haystack.len().
While this is semantically the same as something like
haystack.iter().rposition(|&b| b == needle1 || b == needle2), this
routine will attempt to use highly optimized vector operations that can be
an order of magnitude faster (or more).
§Example
This shows how to find the last position of one of two possible bytes in a haystack.
use memchr::memrchr2;
let haystack = b"the quick brown fox";
assert_eq!(memrchr2(b'k', b'o', haystack), Some(17));