Hello everybody,
I am to new to this list (and to the very promising language that Rust
definitely is).
While learning Rust and writing my first applications in it I stumbled
upon the problem of matching an owned box to a variable pattern. I am
wondering if there is any syntax to convert an owning pointer to a
borrowed one when binding to a variable pattern.
Please consider the following enum:
enum Foobar {
Foo(~str),
Bar(~str)
}
Now, I would like to write a function
fn f<'a>(fb: &'a Foobar) -> &'a str
that takes a borrowed pointer to Foobar and returns a borrowed pointer
to a string from either Foo or Bar. I know I can write it this way:
fn f<'a>(fb: &'a Foobar) -> &'a str {
let rs = match *fb {
Foo(ref rs) => rs,
Bar(ref rs) => rs
};
let s : &'a str = *rs;
s
}
However, is there an alternative that does not introduce (ref)
indirection? Something that would make s in the code below be of type
&str instead of ~str?
fn f<'a>(fb: &'a Foobar) -> &'a str {
match *fb {
Foo(s) => s,
Bar(s) => s
}
}
Wojciech
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