Junglefowl

一次证明就残酷地谋杀Rust的类型系统。

Junglefowl在Rust类型上运行Peano算术，编译时验证。

为什么？

因此我们可以做理论上很难的事情，比如这些const泛型切片

use junglefowl::*;

// Accept only `u8` arrays with exactly 3 elements:
fn picky<T: Nest<Element = u8, Length = peano!(3)>>(_: &T) {}

// Create an array with 5 elements:
let n12345 = nest![1, 2, 3, 4, 5];

// Split it after its second element without changing anything in memory:
let (left, right) = split!(n12345, 2);

// And we can prove that the second segment will have exactly two elements:
picky(&right);
// picky(&left); // won't compile!

// And know exactly what its elements are:
assert_eq!(nest![3, 4, 5], right);

怎么做？

这是我们Peano编码

0 <-->                ()
1 <-->           ((), ())
2 <-->      ((), ((), ()))
3 <--> ((), ((), ((), ())))

注意，多亏了Rust语法的巧妙滥用，这些既是类型也是值。

接下来有一个宏，让你忘记你刚才读到的内容

peano!(0);
 --> ()
peano!(42);
 --> ((), ((), ((), ((), ((), ((), ((), ((), ((), ...)))))))))

注意，这个宏展开为类型，所以你会这样使用它

let x: peano!(42) = todo!();

而不是这样

let x = peano!(42); // bad!

接下来还有一大堆其他的东西，但为了不解释所有这些，请看这个编译过程

static_assert_eq!(peano!(39), sub!(peano!(42), peano!(3)));

展开为

enum False {} // uninstantiable type

//      this part vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv evaluates to zero when the two sides are equal
const _: [False; (peano!(39) != sub!(peano!(42), peano!(3))) as usize] = [];
// ... which makes the list length zero, which matches the right-hand side (and couldn't be nonzero since its members are uninstantiable)
// learned the list length trick from the `static_assertions` crate, so all credit there!

展开上面的有趣部分（并反转，以便将!=变成==）

peano!(39) == <         peano!(42)          as peano::Sub<     peano!(3)     >>::Difference;
peano!(39) == <((), ((), ((), peano!(39)))) as peano::Sub<((), ((), ((), ())))>::Difference;

pub trait Sub<R: peano::N>: peano::N { type Difference: peano::N; } // sealed trait
impl<T: peano::N> Sub<()> for T { type Difference = Self; } // subtracting zero is our super-simple base case
impl<L: peano::N + Sub<R>, R: peano::N> Sub<((), R)> for ((), L) { type Difference = sub!(L, R); } // otherwise, reduce the problem until it's dividing by zero

peano!(39) == <((), ((), ((), peano!(39)))) as peano::Sub<((), ((), ((), ())))>::Difference;
peano!(39) == <     ((), ((), peano!(39)))  as peano::Sub<     ((), ((), ())) >::Difference;
peano!(39) == <          ((), peano!(39))   as peano::Sub<          ((), ())  >::Difference;
peano!(39) == <               peano!(39)    as peano::Sub<               ()   >::Difference;
peano!(39) ==                 peano!(39)                                                   ;