It's a pattern match, "destructuring" something of type &T
. That is, in
let &x = &1i;
x
has type int
, and value 1. So it's actually the opposite to ref
(which does what @KerrekSB is saying, ref x
captures by reference rather than by value).
One can regard it as similar to
match returns_an_option() {
Some(a) => { ... }
None => { ... }
}
except the constructor of &T
is &
, not Some
or None
.
In this specific instance, I guess seps
is a vector (the error you state indicates it's probably a &[&str]
), so the .iter()
returns an object that implements Iterator<& &str>
, that is, it is an iterator over references to the elements of the vector (&str
), thus, you need to dereference next
somehow to get to the original &str
. This can be done by &
in a pattern match (as the code demonstrates) or with *next
when it is used.
(Note that the &
patterns only work with implicitly copyable types, as one cannot move ownership out from a reference/borrowed pointer (i.e. &T
).)
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