pub fn memrchr2(needle1: u8, needle2: u8, haystack: &[u8]) -> Option<usize>Expand description
Like memrchr, but searches for either of two bytes instead of just one.
This returns the index corresponding to the last occurrence of needle1 or
the last occurrence of needle2 in haystack (whichever occurs later), or
None if neither one is found. If an index is returned, it is guaranteed
to be less than usize::MAX.
While this is operationally the same as something like
haystack.iter().rposition(|&b| b == needle1 || b == needle2), memrchr2
will use a highly optimized routine that can be up to an order of magnitude
faster in some cases.
Example
This shows how to find the last position of either of two bytes in a byte string.
use memchr::memrchr2;
let haystack = b"the quick brown fox";
assert_eq!(memrchr2(b'k', b'q', haystack), Some(8));