key: i64) {// Doesn't compile with Rust 2018:self.map.entry(key).or_insert_with(|| self.def.clone());// | --- --- ^^ --- second borrow occurs...// | | | |// | | | immutable borrow occurs here// | | mutable borrow later used...
| immutable borrow occurs here | mutable borrow later used here For more information about this error, try `rustc --explain E0502`. error: could not compile `hello-world` (bin "hello-world") due to 1 previous error 也就是list.get_interface()这里已经有mutable borrow了。为啥这里NLL不起作...
4 | let s2: &String = &s; | -- immutable borrow occurs here 5 | 6 | s.push('A'); | ^^^ mutable borrow occurs here 7 | 8 | let len = s2.len(); | -- immutable borrow later used here 为什么在那种情况下 Rust 不能保证所有权机制呢?或者是,利用裸指针突破所有权机制,会造成什么...
error[E0502]: cannot borrow `s` as mutable because it is also borrowed as immutable --> src/main.rs:18:5 | 16 | let word = first_word(&s); | -- immutable borrow occurs here 17 | 18 | s.clear(); // error! | ^^^ mutable borrow occurs here 19 | 20 | println!("the firs...
-- immutable borrow occurs here 5 |letr2 = &s; // no problem 6 |letr3 = &mut s; // BIG PROBLEM ^^^ mutable borrow occurs here 7 | 8 | println!("{}, {}, and {}", r1, r2, r3); -- immutable borrow later used here Formore...
我刚从Rust开始。关于这段代码,我在理解可变和不可变借用方面遇到了一些困难。这是我正在测试的代码。我的问题是,为什么编译器不将其标记为这样的错误?immutable borrow occurs here | mutable borrow later used here 赋值的右边不是一个不可变的借来吗为什么只将第一个代码片段标记为错误 ...
// | | | immutable borrow occurs here // | | mutable borrow later used by call // | mutable borrow occurs here } } 然而,如果我们内联or_insert_with的定义和lambda函数,编译器最终可以看到借用规则成立 复制 struct S { map: HashMap, def: String } impl S { fn ...
| ^^ immutable borrow occurs here 5 | dbg!(b, c); | - mutable borrow later used here 可变借用出现后立即重新借用为不可变引用,然后可变引用自身析构。为什么Rust会认为这个不可变的重新借用仍具有可变引用的独占生命周期?虽然上面这个例子没什么问题,但允许将可变引用降级为共享引用实际上引入了潜在的内存安...
| --- mutable borrow occurs here 4 | s2.push_str(" rust"); 5 | println!("{}", s1); | ^^ immutable borrow occurs here 6 | println!("{}", s2); | -- mutable borrow later used here | = note: this error originates in the macro `$crate::format_args_nl` which comes from...
| ^^^ immutable borrow occurs here 我们再次审视我们的Rust代码。诚然,它同时出现了&mut T和&T,但我们的代码确实没有数据争用。我们可以注意到,从我们第一次修改slow的时候起,我们就再也没有使用过fast。但是Rust是不会在乎的。它只知道我们没有遵守引用限制。 另一...