Is there a way to simplify the below code by dereferencing foo
and bar
on the same line as we destructure the tuple?
let a = "a";
let b = "b";
let c = (&a, &b);
// Can we dereference here?
let (foo, bar) = c;
let foo_loc = *foo;
let bar_loc = *bar;
println!("{}", foo_loc);
println!("{}", bar_loc);
You can pattern match the references:
fn main() {
let a = "a";
let b = "b";
let c = (&a, &b);
let (&foo, &bar) = c;
println!("{}", foo);
println!("{}", bar);
}