首页 > 解决方案 > 为什么 rustc 说推断的类型不匹配?

问题描述

如果我尝试创建一个函数指针向量,编译器总是会抱怨错误的类型(尽管我没有明确声明任何类型):

fn abc() {}

fn def() {}

fn main()
{
    let a = vec![abc, def];
}

https://play.rust-lang.org/?version=stable&mode=debug&edition=2018&gist=ef52fe778db1112433d20dd02b3d0b54

error[E0308]: mismatched types
 --> src/main.rs:5:13
  |
5 |     let a = vec![abc, def];
  |             ^^^^^^^^^^^^^^ expected slice, found array of 2 elements
  |
  = note: expected struct `Box<[fn() {abc}], _>`
             found struct `Box<[fn(); 2], std::alloc::Global>`
  = note: this error originates in a macro (in Nightly builds, run with -Z macro-backtrace for more info)

error: aborting due to previous error

For more information about this error, try `rustc --explain E0308`.
error: could not compile `tes`

To learn more, run the command again with --verbose.

标签: typesrusttypeerrortype-inference

解决方案


我相信 Rust 试图为函数赋予唯一的类型,所以只需指定它是一个函数指针 Vec。这也避免了额外的间接和分配Box<dyn Fn()>

fn abc() {}
fn def() {}

fn main() {
    let a: Vec<fn()> = vec![abc, def];
}

或者,可以推断出“cast”第一个和其余部分:

fn abc() {}
fn def() {}

fn main() {
    let a = vec![abc as fn(), def];
}

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