鏽表達 「類型不匹配：預計`（）`」

Question

我試圖實現鏽fizzbuzz，並與一些神祕的錯誤失敗：

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; 
    } 
} 

錯誤：

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 
       }); 

或者在較新版本鏽用稍微修改的錯誤消息：

error: if may be missing an else clause [--explain E0308] 
        if i % 3 == 0 { "Fizz" }, 
        ^^^^^^^^^^^^^^^^^^^^^^^^ expected(), found &-ptr 
note: in this expansion of format_args! 
note: in this expansion of print! (defined in <std macros>) 
note: in this expansion of println! (defined in <std macros>) 
note: expected type `()` 
note: found type `&'static str` 

error: if may be missing an else clause [--explain E0308] 
       if i % 5 == 0 { "Buzz" }, 
       ^^^^^^^^^^^^^^^^^^^^^^^^ expected(), found &-ptr 
note: in this expansion of format_args! 
note: in this expansion of print! (defined in <std macros>) 
note: in this expansion of println! (defined in <std macros>) 
note: expected type `()` 
note: found type `&'static str` 

error: if may be missing an else clause [--explain E0308] 
       if !(i % 3 == 0 || i % 5 == 0) { 
       ^expected(), found integral variable 
note: in this expansion of format_args! 
note: in this expansion of print! (defined in <std macros>) 
note: in this expansion of println! (defined in <std macros>) 
note: expected type `()` 
note: found type `_` 

我發現why does removing return give me an error: expected '()' but found，但添加return建議沒有幫助。

這些錯誤意味着什麼，我將來如何避免它們？

Answer 1

問題是if i % 3 == 0 { "Fizz" }返回單位()或&'static str。在兩種情況下更改if表達式以返回相同類型，例如，通過添加else { "" }。

Answer 2

就我而言，作爲初學者，我加了「;」在return聲明

fn add_one(x: i32) -> i32 { 
    x + 1; // remove ";" from return 
}