阴影变量绑定的资源是否立即释放？

Question

根据Rust的书,"当一个绑定超出范围时,它们被绑定的资源被释放".这也适用于阴影吗？

例:

fn foo() {
    let v = vec![1, 2, 3];
    // ... Some stuff
    let v = vec![4, 5, 6]; // Is the above vector freed here?
    // ... More stuff
} // Or here?

Answer 1

不,它不会立即释放.让Rust 告诉我们自己:

struct Foo(u8);

impl Drop for Foo {
    fn drop(&mut self) { println!("Dropping {}", self.0) }
}

fn main() {
    let a = Foo(1);
    let a = Foo(2);

    println!("All done!");
}

输出是:

All done!
Dropping 2
Dropping 1

对我来说,在将变量转换为某种引用但不关心原始变量的情况下,这已派上用场了.例如:

fn main() {
    let a = Foo(1);
    let a = &a; 
}