为什么间接级别允许省略“mut”？

Question

为什么write1 的参数不需要设置为 mut？它将值直接传递给另一个需要mut 访问权限的函数。

在write1 中需要mut 的任何理由似乎都适用于write2，但编译器不同意：

use std::io::Write;

fn main() {
    let out = &mut std::io::stdout();
    write1(out)
}

fn write1(mut out: impl Write) {      // `mut` not required. Adding `mut` produces warning: "variable does not need to be mutable"
    write2(out)
}

fn write2(mut out: impl Write) {  // `mut` required! Otherwise error "cannot borrow `out` as mutable, as it is not declared as mutable"
    writeln!(out, "hi").unwrap();
}

playground link

Answer 1

在write1 中，您并没有改变out，而是将它（移动）给write2，就像main 将它移动到write1 一样。

write2 可能像这样更清晰：

fn write2(out: impl Write) {
    let mut out = out;
    writeln!(out, "hi").unwrap();
}

引用会非常不同。在这种情况下，您可以按照直觉传播 mut：

fn main() {
    let mut out = &mut std::io::stdout();
    write1(&mut out)
}

fn write1(out: &mut impl Write) {
    write2(out)
}

fn write2(out: &mut impl Write) {
    writeln!(out, "hi").unwrap();
}