error[E0596]: cannot borrow `x` as mutable, as it is not declared as mutable --> src/main.rs:4:9 | 3 |letx = 5; | -help: consider changing this to be mutable: `mut x` 4 |lety = &mut x; | ^^^ cannot borrow as mutable 左右滑动查看完整代码 3.借用规则 有人可能会有疑问,...
error[E0596]: cannot borrow `x` as mutable, as it is not declared as mutable --> src/main.rs:4:9 | 3 | let x = 5; | - help: consider changing this to be mutable: `mut x` 4 | let y = &mut x; | ^^^ cannot borrow as mutable 1. 2. 3. 4. 5. 6. 7. 3.借用规则...
("{:?}", r1); // 错误 Cannot borrow `r1` as immutable because it is also borrowed as mutable println!("{}", r2); } println!("{}", r1); 你可能会好奇,明明传入方法的是引用类型,为什么前两条报错信息中会显示 *r1?这是因为发生了隐式重借用,r1.len() 实际上是 String::len(&*r1...
let x = 5; let y = &x; let z = &mut x; // Error: Cannot borrow `x` as mutable because it is also borrowed as immutable 不允许在不可变引用存在时创建可变引用。 没有标记可变的闭包变量 let mut count = 0; let mut inc = || { count += 1; // Error: Cannot borrow immutable...
error[E0499]: cannot borrow `*self` as mutable more than once at a time 虽然这一类错误长得一样,但是我这里的错误可能并不是大家常遇到的那些妖艳错误,废话不多说,一起来看看。 重构前的正确代码 struct Test { a : u32, b : u32 } impl Test { fn increase(&mut self) { let mut a = ...
^^^ cannot borrowasmutable help:trait`DerefMut` is required to modify through a dereference, but it is not implementedfor`Pin<Box<Foo>>` 我们暂且不要去理解这个错误,可以认为是编译器阻止我们创建&mut Foo类型变量,即基本事实 3 所说的内容。当我们把mut关键字去掉时,如下: fnmain() {letbox...
【Rust问答】借用值的使用是否会影响借用检查的结果,根据借用检查规则,以下代码会报错letmutv=vec![1,2,3,4,5];letthird=&v[0];v.push(6);//cannotborrow`v`asmutablebecauseitisalsoborrowedasimmutableprintln!("thirdis{}",third);但是为什么将最后一行去掉之后,
borrow和mutable都是新的概念。对于新的概念,我们会习惯地用熟知的知识去类比。如果套用函数式编程中不可变的特性,大体可以猜到 Rust 中的变量默认是不可变的。但是 cannot borrow as mutable 中borrow 确实是有点超出认知范围。那么此时弄清定义是非常有必要的。澄清概念...
error[E0596]: cannot borrow `*some_string` as mutable, as it is behind a `&` reference--> src/main.rs:8:57 | fn change(some_string: &String) {--- help: consider changing this to be a mutable reference: `&mut String`8 | some_string.push_str(", world");^^^ ...
不能在拥有不可变引用的同时创建可变引用。编译时会报出cannot borrowsas immutable because it is also borrowed as mutable错误。 fnmain(){letmuts=String::from("hello");letr1=&s;// 没问题letr2=&s;// 没问题letr3=&muts;// 错误println!("{}",r2)} ...