将枚举变量用作函数的奇怪语法是什么? [英] What is this strange syntax where an enum variant is used as a function?
问题描述
Below is the example given by the mod documentation of syn::parse
.
enum Item {
Struct(ItemStruct),
Enum(ItemEnum),
}
struct ItemStruct {
struct_token: Token![struct],
ident: Ident,
brace_token: token::Brace,
fields: Punctuated<Field, Token![,]>,
}
impl Parse for Item {
fn parse(input: ParseStream) -> Result<Self> {
let lookahead = input.lookahead1();
if lookahead.peek(Token![struct]) {
input.parse().map(Item::Struct) // <-- here
} else if lookahead.peek(Token![enum]) {
input.parse().map(Item::Enum) // <-- and here
} else {
Err(lookahead.error())
}
}
}
input.parse().map(Item::Struct)
是一个有效的普通 Rust 语法(看起来不像 Item::Struct
不是一个函数),或者它是 proc_macro
库的一种特殊语法?如果是后者,是否有proc_macro
特定语法规则的文档?
Is input.parse().map(Item::Struct)
a valid normal Rust syntax (appears not as Item::Struct
is not a function), or is it a kind of special syntax for proc_macro
libs? If the latter is the case, is there a documentation of the proc_macro
specific syntax rules?
推荐答案
此语法是标准的 Rust 语法.您可以使用元组结构或类似元组结构的枚举变体作为函数.请看这个小例子:
This syntax is standard Rust syntax. You can use tuple struct or tuple struct-like enum variants as functions. See this small example:
enum Color {
Str(String),
Rgb(u8, u8, u8),
}
struct Foo(bool);
// Use as function pointers (type annotations not necessary)
let f: fn(String) -> Color = Color::Str;
let g: fn(u8, u8, u8) -> Color = Color::Rgb;
let h: fn(bool) -> Foo = Foo;
在下一个示例中,这些函数被直接传递给另一个函数(如 Option::map
)(Playground):
In the next example, those functions are directly passed to another function (like Option::map
) (Playground):
// A function which takes a function
fn string_fn<O, F>(f: F) -> O
where
F: FnOnce(String) -> O,
{
f("peter".to_string())
}
string_fn(|s| println!("{}", s)); // using a clojure
string_fn(std::mem::drop); // using a function pointer
// Using the enum variant as function
let _: Color = string_fn(Color::Str);
您可以在 本书的这一章.
这篇关于将枚举变量用作函数的奇怪语法是什么?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!