为什么 Rust 不推断此闭包的返回类型？

Question

The Rust compiler is usually able to infer the type of an expression that is returned from a closure:

fn main() {
    let a_closure = |num|{
        num+1.0
    };
    println!("{}", a_closure(1.0));
}

But the compiler is unable to infer the type when I define the same closure using a return statement :

fn main() {
    let a_closure = |num|{
        return num+1.0
    };
    println!("{}", a_closure(1.0));
}

/*
    error[E0308]: mismatched types
     --> src/main.rs:3:9
      |
    3 |         return num+1.0
      |         ^^^^^^^^^^^^^^ expected `()`, found `f64`
*/

I'm surprised that Rust can't infer the type here: is it possible to use a return statement in a closure without preventing the compiler from inferring its return type?

Answer 1

This is due to the lack of a semicolon. When you have no semicolon, the last expression is returned, and this last expression is return num + 1.0 . Since a return statement makes the program jump somewhere, else, it's value can be anything, for example:

fn main() {
    let a: String = return;
}

However, if the compiler sees no direct type assigned to it, it will pick the type () as the value of the return statement.

So what happened is:

1. The compiler sees the last expression in the closure, and assigns it the type () by default.
2. The compiler then sees an explicit return inside the closure, which returns the type i32 .

So since there are two attempts to return from the closure, and they each return different types, that's a type mismatch.