rust - 当引用不需要时，为什么原始指针需要 const 关键字？

Question

对于已注释掉的行，请引用下面的代码。它不能被 Rust 编译器编译，这意味着我们需要指定像 *const i32 这样的常量。显式用于 Rust 中的原始指针。但是对于对其指针对象不可变的引用，我们不需要像使用原始指针那样显式指定常量。那么，为什么 Rust 不使用更对称的方式来表达引用和原始指针的相同含义，因为引用实际上是引擎盖下的指针？

fn main() {
    let x1: &i32;
    let x2: &mut i32;
    // let p1: *i32;  // This line doesn't compile!!!
    let p1: *const i32;  // we need to specify the constness explicitly.
    let p2: *mut i32;
}

Answer 1

早期版本的 Rust(1.0 之前)使用 *T为 const原始指针(与您注意的语法相同，今天不起作用)。 RFC 0068更名 *T至 *const T , 理由是使用 *T 更容易出错从现有 C 代码转录类型签名时:

The current difference in Rust unsafe pointers types with C pointers types is proving to be too error prone to realistically enable these optimizations at a future date. By renaming Rust's unsafe pointers to closely match their C brethren, the likelihood for erroneously transcribing a signature is diminished.

RFC 还指出(正如您所做的那样)，这会导致引用和原始指针之间的命名不一致:

Today's unsafe pointers design is consistent with the borrowed pointers types in Rust, using the mut qualifier for a mutable pointer, and no qualifier for an "immutable" pointer. Renaming the pointers would be divergence from this consistency, and would also introduce a keyword that is not used elsewhere in the language, const.