Optimizing a Math Expression Parser in Rust 1 year ago
`n` is the same type as `s` from "match s" and 'n' is just `s` but renamed, if none of the previous conditions passed.
Because `match <exp>` could have contained an expression, you might need to handle a "catch all" case where you can refer to the result of that expression.
The code could have been `match s.doSomething() { ...`. The lines above what you have quoted just compare the result to a couple of a constants. If none are true, the line that you have quoted is equivalent to renaming the result of that expression to `n` and then handling that case.