rust - "mismatched types: expected ` ( )`"在使用 if 表达式时意味着什么？

Question

我尝试在 Rust 中实现 fizzbuzz，但由于一些神秘错误而失败了:

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;
    }
}

错误:

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
                 });

较新版本的 Rust 有一个稍微修改过的错误消息:

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}`

我找到了 why does removing return give me an error: expected '()' but found ，但按照建议添加 return 没有帮助。

这些错误是什么意思，以后我该如何避免这些错误？

Answer 1

问题是 if i % 3 == 0 { "Fizz"} 返回单元 () 或 &'static str .更改 if 表达式以在两种情况下返回相同的类型，例如通过添加 else { ""} 。