Is core::mem::replace(map.get_mut(&key), src)
same as map.entry(key).and_modify(move |dest| core::mem::replace(dest, src))
?
map.get_mut(&key)
returns an Option
, so the first snippet of code doesn't compile.
ignore it, just thinking about the &mut T
in it.
Yes, of course. A &mut T
is a &mut T
, regardless of where you got it from.