函数的Rust HashMap的类型签名

问题描述

我创建了一个HashMap，它将字符串映射到类型为 Vec< Expression>的函数。 ->表达式，其中表达式是我定义的类型。有问题的代码是：

I create a HashMap which maps strings to functions of type Vec<Expression> -> Expression, where Expression is a type I have defined. The code in question is:

let functions: HashMap<_, _> = vec!(("+", Box::new(plus))).into_iter().collect();

如果我让Rust为我推断出类型，如上面的代码所示，它将编译并运行良好，如上面的代码所示。但是，如果我尝试指定类型，则无法编译：

If I let Rust infer the type for me, as in the code above, it compiles and runs fine, as in the code above. However, if I try to specify the type, it doesn't compile:

let functions: HashMap<&str, Box<Fn(Vec<Expression>) -> Expression>> =
    vec!(("+", Box::new(plus))).into_iter().collect();

编译器错误消息不是很有帮助：

The compiler error message isn't very helpful:

let functions: HashMap<&str, Box<Fn(Vec<Expression>) -> Expression>> = vec!(("+", Box::new(plus))).into_iter().collect();
^^^^^^^ a collection of type `std::collections::HashMap<&str, std::boxed::Box<std::ops::Fn(std::vec::Vec<Expression>) -> Expression>>` cannot be built from an iterator over elements of type `(&str, std::boxed::Box<fn(std::vec::Vec<Expression>) -> Expression {plus}>)`

此 HashMap的实际类型是什么？

推荐答案

如果仔细研究一下差异，尽管会令人困惑，但您会得到答案。

If you look closely at the difference you will have your answer, although it can be puzzling.

我希望 plus 被声明为：

fn plus(v: Vec<Expression>) -> Expression;

在这种情况下，加的类型是 fn（Vec< Expression>）->表达式{plus} ，实际上是：它不能是

In this case, the type of plus is fn(Vec<Expression>) -> Expression {plus}, and is actually a Voldemort Type: it cannot be named.

最值得注意的是，它与最终的 fn（Vec< Expression>）->不同。表达式{multiply} 。

Most notably, it differs from an eventual fn(Vec<Expression>) -> Expression {multiply}.

这两种类型可被强制转换为裸 fn（Vec< Expression>）- >表达式（不包括 {plus} / {multiply} 面额）。

Those two types can be coerced into a bare fn(Vec<Expression>) -> Expression (without the {plus}/{multiply} denomination).

这后一种类型可以转换为 Fn（Vec< Expression>）->表达式，这是任何不修改其环境的可调用对象的特征（例如，闭包 | v：Vec< Expression> | v [0] .clone（））。

And this latter type can be transformed into a Fn(Vec<Expression>) -> Expression, which is a trait for any callable which do not modify their environments (such as the closure |v: Vec<Expression>| v[0].clone()).

但是问题是，当 fn（ a）-> b {plus} 可以转换为 fn（a）-> b 可以转换为 Fn（a）-> b ...转换需要更改内存表示形式。这是因为：

The problem, however, is that while fn(a) -> b {plus} can be transformed into fn(a) -> b which can be transformed into Fn(a) -> b... the transformation requires a change of memory representation. This is because:

• fn（a）-> b {plus} 是零大小的类型，


• fn（a）-> b 是函数的指针，


• Box< Fn（a）-> b> 是一个装箱的特征对象，通常表示虚拟指针和数据指针。

• fn(a) -> b {plus} is a zero-sized type,
• fn(a) -> b is a pointer to function,
• Box<Fn(a) -> b> is a boxed trait object which generally means both a virtual pointer and a data pointer.

因此类型归属不起作用，因为它只能执行免费的强制转换。

And therefore the type ascription doesn't work, because it can only perform cost-free coercions.

解决方案是在为时已晚之前执行转换：

The solution is to perform the transformation before it's too late:

// Not strictly necessary, but it does make code shorter.
type FnExpr = Box<Fn(Vec<Expression>) -> Expression>;

let functions: HashMap<_, _> =
    vec!(("+", Box::new(plus) as FnExpr)).into_iter().collect();
               ^~~~~~~~~~~~~~~~~~~~~~~~

或者也许您宁愿保留未装箱的功能：

Or maybe you'd rather keep unboxed functions:

// Simple functions only
type FnExpr = fn(Vec<Expression>) -> Expression;

let functions: HashMap<_, _> =
    vec!(("+", plus as FnExpr)).into_iter().collect();

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