Blocks Have Values in Rust in Rust

In Rust, almost everything is an expression — a piece of code that produces a value. Even a whole { ... } block is an expression: it evaluates to its last expression, as long as that last expression has no semicolon.

let y = {
    let a = 3;
    a + 1       // no semicolon: this value flows out of the block
};
// y is now 4

Functions work exactly the same way: a function's body is a block, so a function returns its last expression automatically. You only need the return keyword for early exits.

fn add_one(n: i32) -> i32 {
    n + 1       // this is the return value — no `return`, no semicolon
}

The trap: one accidental semicolon

A semicolon turns an expression into a statement, and a statement's value is () — the empty "unit" type. So writing n * n; as the last line makes the function return () instead of a number, and the compiler stops you:

error[E0308]: mismatched types
 --> trap.rs:1:22
  |
1 | fn square(n: i32) -> i32 {
  |    ------            ^^^ expected `i32`, found `()`
  |    |
  |    implicitly returns `()` as its body has no tail or `return` expression
2 |     n * n;
  |          - help: remove this semicolon to return this value

error: aborting due to 1 previous error

Read the help line — the compiler even tells you the fix: remove this semicolon to return this value. Every Rust programmer hits this error; now you know exactly what it means.

Assignment
Follow the Coding Tutorial:

• Define a function square(n: i32) -> i32 whose body is just n * n — no return, no semicolon.
• Bind doubled to a block that declares let x = 21; and ends with the expression x * 2.
• Print both results.

The expected output is:

doubled = 42
square(6) = 36

Hint
If you see expected `i32`, found `()`, look for a semicolon on the line that should produce the value.