Why does an if without an else always result in () as the value?

Question

From this tutorial:

An if without an else always results in () as the value.

Why does Rust impose this restriction and doesn\'

Answer 1

For your example, the right question is: “What would the value of y be if x is not 5?”. What would happen here?

let x = 3;
let y = if x == 5 { 10 };
println!("{}", y);  // what?!

You could think that the if-without-else-expression could return an Option<_>, but...

1. this would mean that the core language depends on yet another library item (those are then called lang items) which everyone tries to avoid
2. you don't run into this situation too often
3. you can get the same behavior by adding only a little bit of code (Some() & else { None })

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In Rust, nearly everything is an expression (with the exception of let-bindings and expressions ending with a semicolon, so called expression statements). And there are a few examples of expressions always returning (), because nothing else makes sense. These include (compound-)assignments (why?), loops and if-without-else.