我编写了这段代码,该代码多次借用了一个可变变量,并且编译时没有任何错误,但是根据The Rust Programming Language的说法,该代码不应编译:
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');
}
(playground)
这是Bug还是Rust中有新功能?
最佳答案
这里没有任何奇怪的事情发生。每当test_three
函数结束其工作时(即在调用之后),可变的借位就变为无效:
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
}
该函数不保留其参数-它仅对它指向的
String
进行突变,然后立即释放借用,因为不再需要它: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