in programming, rust

Take this example of Rust’s if let syntax:

if let Some(x) = my_fn() {

Many explanations I’ve seen of this go something like:

To make this easy to understand: if the expression on the right hand side of the = matches Some(x) do something with x inside the if block.

While this is true, I wondered why, if we’re doing pattern matching, there is only one = (instead of, maybe the comparison operator, ==). Overall, this wasn’t cemented for me until thinking about it in the following way:

Matching is destructuring, but for when you don’t know if the destructuring will succeed.

With that in mind, let’s recap on what destructuring is:

fn main() {
    let (x, y, z) = my_tuple_fn();

fn my_tuple_fn() -> (i32, i32, i32) {
    (1, 2, 3)

This evaluates the expression my_tuple_fn() and binds x, y, and z to the result. We’re not left with a tuple, we’ve bound the contents of the tuple to three variable names.

Now imagine we want to do something like this (it won’t compile):

fn main() {
    let Some(x) = my_option_fn();

fn my_option_fn() -> Option<i32> {

The compiler gives us this helpful message:

 cargo run
   Compiling if_let v0.1.0 (/Users/jr/code/learning/if_let)
error[E0005]: refutable pattern in local binding
 --> src/
2 |     let Some(x) = my_option_fn();
  |         ^^^^^^^ pattern `None` not covered
  = note: `let` bindings require an "irrefutable pattern", like a `struct` or an `enum` with only one variant
  = note: for more information, visit
  = note: the matched value is of type `Option<i32>`
help: you might want to use `let else` to handle the variant that isn't matched
2 |     let Some(x) = my_option_fn() else { todo!() };
  |                                  ++++++++++++++++

For more information about this error, try `rustc --explain E0005`.
error: could not compile `if_let` due to previous error

Ok, makes sense. If we can’t be sure that my_option_fn will evaluate to the Some variant, then we need to handle — or ignore — the case where it doesn’t. The compiler is suggesting we can fix this using let else, which is perhaps for another time. For the purposes of this article, though, this experiment really helped me understand why if let is written the way it is. We must try the destructuring and if it succeeds, we can do someting with the value bound to the destructured variable name.