Type inference
- Rust course
- Intermediate
- Lesson 43/115
- 4 min
Learn type inference in closures and when explicit types are required.
Type inference
A closure's parameter and return types are inferred from the first use and then fixed:
let ikile = |x| x * 2;
println!("{}", ikile(5)); // x fixed as i32
// println!("{}", ikile(2.5)); // error: expected i32, found f64Unlike a generic function, a closure works with a single type. Different types need separate closures or a generic function.
Explicit types
Types can be written explicitly for readability or to resolve ambiguity. If the return type is written, the body must be in braces:
let uzunluk = |s: &str| -> usize { s.chars().count() };
println!("{}", uzunluk("kitap")); // 5A closure's type is an anonymous type the compiler generates; even if two closures have the same signature, their types differ.
Practice Inference and explicit type
Write the type-inferred let ikile = |x| x * 2; and the explicitly typed let uzunluk = |s: &str| -> usize { s.chars().count() }; closures. Print the results of ikile(5) and uzunluk("kitap") separated by a space.
Starter code
fn main() {
// buraya yaz
}Sample solution
fn main() {
let ikile = |x| x * 2;
let uzunluk = |s: &str| -> usize { s.chars().count() };
println!("{} {}", ikile(5), uzunluk("kitap"));
}Expected output
10 5
Questions and answers
Review the lesson questions together with their answers. To test yourself, solve the lesson interactively.
1 What happens if <code>let f = |x| x;</code> is never called?
- x is assumed to be i32
- It only warns
- Compile error: the type of x can't be inferred Correct answer
- f becomes generic
Explanation: Without a use there is no type information; in that case write the type explicitly: |x: i32| x.
2 Return type
let uzunluk = |s: &str| ____ { s.chars().count() };let uzunluk = |s: &str| -> usize { s.chars().count() };Explanation: A closure's return type is written with -> and the body needs braces.
3 Code and result
| Term | Meaning |
|---|---|
let f = |x| x * 2; f(5) |
10 |
f(5); f(2.5) |
Type error |
let f = |x| x; (never called) |
Inference error |
|s: &str| -> usize { .. } |
Explicit types |