“不匹配的类型:预期的‘()’"是什么意思?使用 if 表达式时是什么意思? [英] What does "mismatched types: expected `()`" mean when using an if expression?

问题描述

我尝试在 Rust 中实现 fizzbuzz，但由于一些神秘的错误而失败:

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;
    }
}

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 有一个稍微修改的错误消息:

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}`

我发现 ​​为什么删除返回给我有一个错误:预期的 '()' 但找到了，但是按照建议添加 return 没有帮助.

I found why does removing return give me an error: expected '()' but found, but adding return as suggested didn't help.

这些错误是什么意思，我将来如何避免它们?

What do these errors mean and how do I avoid them in the future?

推荐答案

问题在于 if i % 3 == 0 { "Fizz" } 返回任一单元 () 或 &'static str.更改 if 表达式以在两种情况下返回相同的类型，例如添加 else { "" }.

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 { "" }.

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