rust - 变量不需要是可变的，但它确实是可变的

Question

我有一个可以工作的宏:

#[macro_export]
macro_rules! btreemap {
    ($( $key: expr => $val: expr ),*) => {{
         let mut map = ::std::collections::BTreeMap::new();
         $( map.insert($key, $val); )*
         map
    }}
}

但编译器警告:

warning: variable does not need to be mutable
  --> src/util.rs:16:11
   |
16 |          let mut map = ::std::collections::BTreeMap::new();
   |              ----^^^
   |              |
   |              help: remove this `mut`
   | 
  ::: src/foo.rs:49:13
   |
79 |                 foo: btreemap!{},
   |                      ----------- in this macro invocation
   |
   = note: this warning originates in a macro (in Nightly builds, run with -Z macro-backtrace for more info)

如果我删除 mut ，我收到一个错误:

error[E0596]: cannot borrow `map` as mutable, as it is not declared as mutable
   --> src/util.rs:17:10
    |
16  |          let map = ::std::collections::BTreeMap::new();
    |              --- help: consider changing this to be mutable: `mut map`
17  |          $( map.insert($key, $val); )*
    |             ^^^ cannot borrow as mutable
    | 
   ::: src/bar.rs:110:18
    |
116 |         let bar = btreemap!{quux => wibble};
    |                   -------------------------- in this macro invocation
    |
    = note: this error originates in a macro (in Nightly builds, run with -Z macro-backtrace for more info)

即如果 mut它警告空调用，但如果不是非空调用错误。
我怎样才能解决这个问题？

Answer 1

How can I fix this?

您可以通过添加 #[allow(unused_mut)] 来抑制警告。到宏生成的代码:

macro_rules! btreemap {
    ($( $key: expr => $val: expr ),*) => {{
        #[allow(unused_mut)]
        let mut map = ::std::collections::BTreeMap::new();
        $( map.insert($key, $val); )*
        map
    }}
}

警告与警告所能获得的一样良性，这种警告抑制经常出现在宏定义中。
另一种可能性是你已经发现的，特殊情况下的空扩展。但我更喜欢显式警告抑制，因为它消除了(虚假)警告而不会使宏复杂化。您介绍的特殊情况实际上并不是必需的 - 没有它会生成完全相同的代码。