错误[E0507]:无法移出借来的内容 [英] error[E0507]: Cannot move out of borrowed content
问题描述
我正在尝试在 Rust 中创建一个词法分析器，同时对它相对较新，但具有 C/C++ 背景.我在以下代码中遇到了 Rust 如何分配内存的问题，这会生成错误无法移出借用的内容".我已经阅读了 cargo --explain E0507 ，其中详细介绍了可能的解决方案，但我正在努力掌握 Rust 和 C/C++ 如何管理内存之间的潜在差异.本质上，我想了解如何在 Rust 中管理动态内存(或者更好的方式来实现我正在做的事情).
I'm trying to make a lexer in Rust while being relatively new to it but with a background in C/C++. I'm having problems with how Rust allocates memory in the following code, which generates the error "Cannot move out of borrowed content". I've read cargo --explain E0507 which details possible solutions, but I'm struggling to grasp the underlying differences between how Rust and C/C++ manage memory. In essence, I want to understand how to manage dynamic memory in Rust (or a better way to achieve what I'm doing).
错误是:
error[E0507]: cannot move out of borrowed content
--> <anon>:65:16
|
65 | return self.read_tok.unwrap();
| ^^^^ cannot move out of borrowed content
error[E0507]: cannot move out of borrowed content
--> <anon>:73:16
|
73 | return self.peek_tok.unwrap();
| ^^^^ cannot move out of borrowed content
error: aborting due to 2 previous errors
代码是:
use std::fmt;
#[derive(Debug, PartialEq)]
pub enum TokenType {
EndOfFile,
Illegal
}
pub struct Token {
token_type: TokenType,
value: String
}
impl Token {
pub fn new(token_type: TokenType, value: String) -> Token {
return Token {
token_type: token_type,
value: value
};
}
pub fn is_token_type(&self, token_type: TokenType) -> bool {
return self.token_type == token_type;
}
}
impl fmt::Debug for Token {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
write!(f, "{:?}[{}]", self.token_type, self.value)
}
}
pub struct Lexer {
input: String,
read_pos: usize,
peek_pos: usize,
ch: char,
read_tok: Option<Token>,
peek_tok: Option<Token>
}
const EOF: char = 0 as char;
impl Lexer {
pub fn new(input: &str) -> Lexer {
return Lexer {
input: input.to_string(),
read_pos: 0,
peek_pos: 1,
ch: EOF,
read_tok: None,
peek_tok: None
};
}
pub fn next_token(&mut self) -> Token {
if self.peek_tok.is_none() {
self.read_tok = Some(self.get_next_token());
} else {
self.read_tok = self.peek_tok.take();
}
return self.read_tok.unwrap();
}
pub fn peek_token(&mut self) -> Token {
if self.peek_tok.is_none() {
self.peek_tok = Some(self.get_next_token());
}
return self.peek_tok.unwrap();
}
fn get_next_token(&mut self) -> Token {
let ch = self.next_char();
let tok: Token;
match ch {
EOF => { tok = Token::new(TokenType::EndOfFile, "".to_string()); }
_ => { tok = Token::new(TokenType::Illegal, ch.to_string()); }
}
return tok;
}
fn next_char(&mut self) -> char {
if self.peek_pos >= self.input.len() {
self.ch = EOF;
} else {
self.ch = self.input.chars().nth(self.peek_pos).unwrap();
}
self.read_pos = self.peek_pos;
self.peek_pos += 1;
return self.ch;
}
}
fn main() {
let input = "let x = 5;";
let mut l = Lexer::new(input);
loop {
let t = l.next_token();
println!("{:?}", t);
if t.is_token_type(TokenType::EndOfFile) {
break;
}
}
}
Rust playground link: https://play.rust-lang.org/?gist=bc85fafa35a5cbbd5ac4066aef9e333c&version=stable&backtrace=0https://play.rust-lang.org/?gist=21cba64f53488ee0a9389c0191c47134&version=stable&backtrace=0
我已经设法在 C++ 中翻译了一个工作实现，它可能会提供有关我正在尝试实现的目标的更多信息:
I've managed to translate a working implementation in C++ which might give some more info on what I'm trying to achieve:
#include <string>
#include <iostream>
enum TokenType {
ENDOFFILE,
ILLEGAL
};
class Token {
private:
enum TokenType token_type;
std::string value;
public:
Token(enum TokenType token_type, std::string value)
{
this->token_type = token_type;
this->value = value;
}
bool is_token_type(enum TokenType token_type)
{
return this->token_type == token_type;
}
std::string to_string()
{
std::string tok;
switch (this->token_type) {
case ENDOFFILE:
tok = "EndOfFile";
break;
case ILLEGAL:
tok = "Illegal[" + this->value + "]";
break;
}
return tok;
}
};
class Lexer {
private:
std::string input;
int read_pos;
int peek_pos;
char ch;
Token *read_tok;
Token *peek_tok;
Token *get_next_token() {
char c = this->next_char();
std::string c_str;
Token *t;
c_str.push_back(c);
switch (c) {
case 0:
t = new Token(ENDOFFILE, "");
break;
default:
t = new Token(ILLEGAL, c_str);
}
return t;
}
char next_char()
{
if (this->peek_pos >= this->input.length()) {
this->ch = 0;
} else {
this->ch = input.at(this->peek_pos);
}
this->read_pos = this->peek_pos;
this->peek_pos += 1;
return this->ch;
}
public:
Lexer (std::string input)
{
this->input = input;
this->read_pos = -1;
this->peek_pos = 0;
this->ch = 0;
this->read_tok = NULL;
this->peek_tok = NULL;
}
Token *next_token()
{
if (this->read_tok != NULL) {
delete this->read_tok;
}
if (this->peek_tok == NULL) {
this->read_tok = this->get_next_token();
} else {
this->read_tok = this->peek_tok;
this->peek_tok = NULL;
}
return this->read_tok;
}
Token *peek_token()
{
if (this->peek_tok == NULL) {
this->peek_tok = this->get_next_token();
}
return this->peek_tok;
}
};
int main(int argc, char **argv)
{
std::string input = "let x = 5;";
Lexer l = Lexer(input);
while (1) {
Token *t = l.next_token();
std::cout << t->to_string() << std::endl;
if (t->is_token_type(ENDOFFILE)) {
break;
}
}
return 0;
}
推荐答案
您已经非常接近正确，但是您的代码存在两个问题.
You came very close to getting it right, but there are two problems with your code.
首先，正如编译器告诉您的，以下内容是被禁止的:
First, as the compiler tells you, the following is prohibited:
self.read_tok = self.peek_tok;
self.peek_tok = None;
第一行尝试将 Option 对象移出 self.peek_tok.在 Rust 中，对象可以移出变量，但不能移出结构字段或切片下标.这是因为编译器可以在移动后检查变量是否未被使用，以及安排不调用其析构函数.这对于存储在结构域或切片内部的对象是不可能的，至少在不增加每个结构或容器的开销的情况下是不可能的.
The first line attempts to move an Option<Token> object out of self.peek_tok. In Rust, objects can be moved out of variables, but not out of structure fields or slice subscripts. This is because the compiler can check that the variable is not used after the move, as well as arrange that its destructor is not invoked. This is not possible for objects stored in fields of structures or inside slices, at least not without adding overhead to every structure or container.
只要将对象存储在支持移动的中间容器中，就可以将对象移出结构体.幸运的是，Option 就是这样一个容器，它的 take() 方法正是为此目的而设计的:
Moving objects out of structs is possible as long as they are stored in an intermediate container that supports moving. Fortunately, Option is such a container, and its take() method is designed for exactly that purpose:
self.read_tok = self.peek_tok.take()
Option::take() 从选项中移动对象，用 None 替换它，并返回对象.
Option::take() moves the object from the option, replaces it with None, and returns the object.
其次，上述问题解决后，编译器在next_token 和peek_token 的return 语句中抱怨移出借来的内容">，因为它们试图将对象移出 Option.在这里，您可以选择克隆 Token，或使用上述 Option::take() 将其移出选项.克隆方法需要将#[derive(Clone)]添加到TokenType和Token，并将返回值改为:
Second, once the above is fixed, the compiler complains of "moving out of borrowed content" on the return statements of next_token and peek_token, because they attempt to move objects out of the Option. Here you have the choice of cloning the Token, or moving it out of the option using Option::take() as above. The cloning approach requires adding #[derive(Clone)] to TokenType and Token, as well as changing the returns to:
// Use as_ref() to convert Option<Token> to Option<&Token>,
// which is unwrapped and the Token inside cloned
self.read_tok.as_ref().unwrap().clone()
通过这些更改，示例可以编译，尽管它仍然将输入标记为非法.
With these changes, the example compiles, although it still flags the input as illegal.
这篇关于错误[E0507]:无法移出借来的内容的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
