Why does the compiler assume that the value of if let should be `()`?

Question

I have the following code:

use std::collections::HashSet;

fn translate() -> Option {
    None
}

fn main() {
    let mut found = HashSet::new()         


        

Answer 1

According to the Rust Book (emphasis mine):

The value of the expression is the value of the last expression in whichever branch was chosen. An if without an else always results in () as the value.

This gives a constraint on the expression value inside the curly braces.

This is correct since the expression type matches ():

if let Some(_) = some() {
    ()
};

This is correct since there's an else statement (and the types between the branches match):

if let Some(_) = some() {
    true
} else {
    false
};

But this is wrong:

if let Some(_) = some() {
    true
};

This answer was inspired by this comment.

Answer 2

When you omit a function's return type, the function actually returns (). That is,

fn foo() {}

is equivalent to:

fn foo() -> () {}

---

If I add return (); at the end of the function I still get the same error. So I'm not even sure if that has something to do with a function return value.

An if let expression that is used as a statement must return (), unless it is the last expression in the function's body, in which case its type must match the function's return type. Since your if let doesn't have an else clause, its type must be ().