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?
Best Answer
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 { "" }.