为什么不能在表达式中使用...语法？

Question

我正在尝试理解...语法.考虑以下程序:

fn how_many(x: i32) -> &'static str {
    match x {
        0 => "no oranges",
        1 => "an orange",
        2 | 3 => "two or three oranges",
        9...11 => "approximately 10 oranges",
        _ => "few oranges",
    }
}

fn pattern_matching() {
    for x in 0..13 {
        println!("{} : I have {} ", x, how_many(x));
    }
}

fn main() {
    // println!("{:?}", (2...6));
    pattern_matching();
}

在上面的程序中,9...11模式匹配中使用的编译很好.当我尝试使用相同的内容时,println!("{:?}", (2...6));我得到编译错误:

error: `...` syntax cannot be used in expressions
  --> src/main.rs:18:24
   |
18 |     println!("{:?}", (2...6));
   |                        ^^^
   |
   = help: Use `..` if you need an exclusive range (a < b)
   = help: or `..=` if you need an inclusive range (a <= b)

我试图理解为什么不可能在内部使用println.

Answer 1

出于同样的原因,你不能&*$#在表达式中使用:Rust的语法不允许这样做.Rust决定对包含范围使用不同的语法.如果你真的关心完整的细节,你可以阅读有关该问题的所有350多条评论.

该TL; DR:

• 没有人喜欢特定的语法
• ... 太接近于错字了 ..
• ...模式已经稳定,必须永远支持,但它会在某个时候被弃用,并且..=也会优先考虑.

该错误消息告诉您使用相应的语法:

   = help: Use `..` if you need an exclusive range (a < b)
   = help: or `..=` if you need an inclusive range (a <= b)

fn how_many(x: i32) -> &'static str {
    match x {
        0 => "no oranges",
        1 => "an orange",
        2 | 3 => "two or three oranges",
        9..=11 => "approximately 10 oranges",
        _ => "few oranges",
    }
}

fn pattern_matching() {
    for x in 0..13 {
        println!("{} : I have {} ", x, how_many(x));
    }
}

fn main() {
    println!("{:?}", 2..=6);
    pattern_matching();
}