为什么在同一范围内可以进行多个可变借用? [英] Why are multiple mutable borrows possible in the same scope?

问题描述

我写了这段代码，它多次借用一个可变变量并且编译没有任何错误，但是根据 Rust 编程语言这不应该编译:

I wrote this code that borrows a mutable variable more than once and compiles without any error, but according to The Rust Programming Language this should not compile:

fn main() {
    let mut s = String::from("hello");

    println!("{}", s);
    test_three(&mut s);
    println!("{}", s);
    test_three(&mut s);
    println!("{}", s);
}

fn test_three(st: &mut String) {
    st.push('f');
}

(游乐场)

这是一个错误还是 Rust 中有新功能?

Is this a bug or there is new feature in Rust?

推荐答案

这里没有发生任何奇怪的事情；每次 test_three 函数结束其工作(即在它被调用之后)时，可变借用变为无效:

Nothing weird happens here; the mutable borrow becomes void every time the test_three function concludes its work (which is right after it's called):

fn main() {
    let mut s = String::from("hello");

    println!("{}", s); // immutably borrow s and release it
    test_three(&mut s); // mutably borrow s and release it
    println!("{}", s); // immutably borrow s and release it
    test_three(&mut s); // mutably borrow s and release it
    println!("{}", s); // immutably borrow s and release it
}

该函数不保存它的参数 - 它只改变它指向的 String 并在之后立即释放借用，因为不再需要它:

The function doesn't hold its argument - it only mutates the String it points to and releases the borrow right afterwards, because it is not needed anymore:

fn test_three(st: &mut String) { // st is a mutably borrowed String
    st.push('f'); // the String is mutated
} // the borrow claimed by st is released

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