What does "mismatched types: expected `()`" mean when using an if expression?
rust
Solution
The problem is that `if i % 3 == 0 { "Fizz" }` returns either unit `()` or `&'static str`. Change the if expressions to return the same type in both cases, for example by adding a `else { "" }`.
Problem
I tried to implement fizzbuzz in Rust and failed with some arcane error: ``` fn main() { let mut i = 1; while i < 100 { println!( "{}{}{}", if i % 3 == 0 { "Fizz" }, if i % 5 == 0 { "Buzz" }, if !(i % 3 == 0 || i % 5 == 0) { i }, ); i += 1; } } ``` Error: ``` error: mismatched types: expected `()` but found `&'static str` (expected () but found &-ptr) if i % 3 == 0 { "Fizz" }, ^~~~~~~~~~ error: mismatched types: expected `()` but found `&'static str` (expected () but found &-ptr) if i % 5 == 0 { "Buzz" }, ^~~~~~~~~~ error: mismatched types: expected `()` but found `<generic integer #0>` (expected () but found integral variable) if !(i % 3 == 0 || i % 5 == 0) { i }); ``` Newer versions of Rust have a slightly modified error message: ``` error[E0317]: if may be missing an else clause --> src/main.rs:7:13 | 7 | if i % 3 == 0 { "Fizz" }, | ^^^^^^^^^^^^^^^^^^^^^^^^ expected (), found &str | = note: expected type `()` found type `&str` error[E0317]: if may be missing an else clause --> src/main.rs:8:13 | 8 | if i % 5 == 0 { "Buzz" }, | ^^^^^^^^^^^^^^^^^^^^^^^^ expected (), found &str | = note: expected type `()` found type `&str` error[E0317]: if may be missing an else clause --> src/main.rs:9:13 | 9 | if !(i % 3 == 0 || i % 5 == 0) { i }, | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ expected (), found integral variable | = note: expected type `()` found type `{integer}` ``` I found why does removing return give me an error: expected '()' but found, but adding `return` as suggested didn't help. What do these errors mean and how do I avoid them in the future?