generics - 为什么 rust 只允许数组大小的独立常量？

Question

我想使用一个函数来连接两个数组，声明如下:

fn concatenate<const COUNT1: usize, const COUNT2: usize>(a: [i32;COUNT1], b: [i32;COUNT2]) -> [i32;{COUNT1+COUNT2}];

问题在于返回类型。这是具体的错误:

error: generic parameters may not be used in const operations
 --> src\main.rs:4:101
  |
4 | fn concatenate<const COUNT1: usize, const COUNT2: usize>(a: [i32;COUNT1], b: [i32;COUNT2]) -> [i32;{COUNT1+COUNT2}] {
  |                                                                                                     ^^^^^^ cannot perform const operation using `COUNT1`
  |
  = help: const parameters may only be used as standalone arguments, i.e. `COUNT1`

这个函数似乎很容易单态化，我不明白为什么编译器不允许它。使用rust 书只说(两次)不允许，但没有解释原因:

Const parameters can be used anywhere a const item can be used, with the exception that when used in a type or array repeat expression, it must be standalone (as described below).

As a further restriction, const parameters may only appear as a standalone argument inside of a type or array repeat expression.

有谁知道这种模式是如何反对 rust 模型的，因为至少从我的角度来看，它绝对不是实现限制。如果有帮助，这是整个功能:

fn concatenate<const COUNT1: usize, const COUNT2: usize>(a: [i32;COUNT1], b: [i32;COUNT2]) -> [i32;{COUNT1+COUNT2}] {
    let mut output = [0i32;{COUNT1+COUNT2}];
    output.copy_from_slice(
        &a.iter().chain(b.iter()).map(|&item| item).collect::<Vec<i32>>()
    );
    output
}

Answer 1

您需要启用 generic_const_exprs 功能，该功能目前仅在 nightly 中可用(rust 1.63):

#![feature(generic_const_exprs)]

fn concatenate<const COUNT1: usize, const COUNT2: usize>(a: [i32;COUNT1], b: [i32;COUNT2]) -> [i32;{COUNT1+COUNT2}] {
    let mut output = [0i32;{COUNT1+COUNT2}];
    output.copy_from_slice(
        &a.iter().chain(b.iter()).map(|&item| item).collect::<Vec<i32>>()
    );
    output
}

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