如何在没有赋值的情况下声明静态可变变量？ [英] How do I declare a static mutable variable without assignment?

问题描述

我尝试了以下操作

struct mbuf
{
  cacheline: *mut [u64],               // great amount of rows follows below
  // ..........
}
static mut arr: [mbuf; 32];                // Q1 my main aim
// something as before but using Vec;      // Q2 also main aim

fn main() {
 // let static mut arr: [mbuf; 32];        // Q3 also doesn't work
 // static mut arr: [mbuf; 32];            // Q3 also doesn't work
}

并出现错误

src/main.rs:74:29: 74:30 error: expected one of `+` or `=`, found `;`
src/main.rs:74   static mut arr: [mbuf; 32];
                                           ^

Q1，Q2，Q3-有可能吗？如何？

Q1,Q2,Q3 - Is it possible and how?

推荐答案

声明时必须分配静态或常量；

A static or constant must be assigned when declared; they can never be assigned to after that.

静态内容必须是纯文字；它不能有任何函数调用。

A static must be purely literals; it cannot have any function calls.

当前常量必须纯粹是文字，但是当 RFC 911，const fn 已实现，将有可能做更多像您想要的事情。

A constant must at present be purely literals, but when RFC 911, const fn is implemented it will be possible to do things more like how you desire.

在函数内部，您可以拥有 static 或 const 项就像在函数外部一样，并且没有区别-在函数内部放置项目（特征和类型定义，函数，& c。）纯粹是将它们隐藏在外部范围之外。因此，您通常也可以使用 let foo 。

Inside a function, you can have static or const items, just as outside a function, and there is no difference—placing items (trait and type definitions, functions, &c.) inside a function purely hides them from the outside scope. Therefore you typically might as well use let foo.

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