我如何制作格式！从条件表达式返回 &str? [英] How do I make format! return a &str from a conditional expression?

问题描述

我遇到了这个问题，其中 format! 在一个模式中创建了一个临时值，据我所知，该值没有锚定到任何东西.

I happened upon this problem where format! creates a temporary value in a pattern that is not anchored to anything, as far as I understand it.

let x = 42;
let category = match x {
    0...9 => "Between 0 and 9",
    number @ 10 => format!("It's a {}!", number).as_str(),
    _ if x < 0 => "Negative",
    _ => "Something else",
};

println!("{}", category);

在这段代码中，category的类型是一个&str，它通过返回一个像Between 0 and 9"这样的文字来满足.代码>.如果我想使用 as_str() 将匹配的值格式化为切片，则会出现错误:

In this code, the type of category is a &str, which is satisfied by returning a literal like "Between 0 and 9". If I want to format the matched value to a slice using as_str(), then I get an error:

error[E0716]: temporary value dropped while borrowed
 --> src/main.rs:5:24
  |
3 |     let category = match x {
  |         -------- borrow later stored here
4 |         0...9 => "Between 0 and 9",
5 |         number @ 10 => format!("It's a {}!", number).as_str(),
  |                        ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^        - temporary value is freed at the end of this statement
  |                        |
  |                        creates a temporary which is freed while still in use
  |
  = note: consider using a `let` binding to create a longer lived value
  = note: this error originates in a macro outside of the current crate (in Nightly builds, run with -Z external-macro-backtrace for more info)

我读了一些书，发现有类似问题的人，但我似乎找不到任何解决方案.

I have done some reading, and found people with similar problems, but I couldn't seem to find any solution.

一个简单的解决方法是让 category 成为 String 而不是 &str，但我不喜欢这个想法必须将 .to_string() 放在模式中每个文字的末尾，因为它不是那么干净.

A simple workaround would be to have category be a String instead of a &str, but I don't like the idea of having to put .to_string() on the end of every literal in the pattern, as it's not as clean.

有没有办法解决这个问题，或者我只需要解决它?

Is there a way to solve the problem, or do I just need to work around it?

推荐答案

这是 90% 的 以切片形式返回本地字符串(&str)，请参阅其他多种解决方案.

This is 90% a duplicate of Return local String as a slice (&str), see that for multiple other solutions.

还有一种额外的可能性，因为这一切都在一个函数中:您可以为 String 声明一个变量，并且仅在需要分配时才设置它.编译器(斜向)建议:

There's one extra possibility since this is all in one function: You can declare a variable for the String and only set it when you need to allocate. The compiler (obliquely) suggests this:

考虑使用 let 绑定来创建更长寿的值

consider using a let binding to create a longer lived value

fn main() {
    let x = 42;
    let tmp;

    let category = match x {
        0...9 => "Between 0 and 9",
        number @ 10 => {
            tmp = format!("It's a {}!", number);
            &tmp
        }
        _ if x < 0 => "Negative",
        _ => "Something else",
    };

    println!("{}", category);
}

这与使用 Cow 大致相同，只是由编译器处理而不是特定类型.

This is mostly the same as using a Cow, just handled by the compiler instead of a specific type.

另见:

• 如何在不使用时有条件地提供默认引用而不执行不必要的计算?
• 是否允许使用多态变量?

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