锈“预期类型"错误会显示完全相同的不匹配类型 [英] Rust "expected type" error prints mismatched types that are exactly the same

问题描述

夜间生锈:

游乐场

struct Foo<T, F: Fn(&T, &T) -> T> {
    value: T,
    func: F
}

fn main() {
    let lambda = |&x, &y| x + y;
    let foo = Foo {
        value: 5 as i32,
        func: lambda
    };
}

错误消息:

Compiling playground v0.0.1 (/playground)
error[E0308]: mismatched types
 --> src/main.rs:8:15
  |
8 |     let foo = Foo {
  |               ^^^ one type is more general than the other
  |
  = note: expected type `std::ops::FnOnce<(&i32, &i32)>`
             found type `std::ops::FnOnce<(&i32, &i32)>`

请注意，期望的类型和找到的类型是字符相同的字符.为什么错误消息中说一种类型比另一种类型更通用，同时又说它们是同一类型?

Note that the expected type and found type are character for character identical. Why is the error message saying that one type is more general than the other, while also saying that they are the same type?

推荐答案

夜间生锈:

这似乎只是夜间构建中的错误"错误消息.在Rust 1.32(稳定版)中，错误告诉您这是生命周期不匹配:

This appears to be just a "bad" error message in a nightly build. In Rust 1.32 (stable), the errors tell you that this is a lifetime mismatch:

error[E0631]: type mismatch in closure arguments
 --> src/main.rs:8:15
  |
7 |     let lambda = |&x, &y| x + y;
  |                  -------------- found signature of `fn(&_, &_) -> _`
8 |     let foo = Foo {
  |               ^^^ expected signature of `for<'r, 's> fn(&'r i32, &'s i32) -> _`
  |
note: required by `Foo`
 --> src/main.rs:1:1
  |
1 | struct Foo<T, F: Fn(&T, &T) -> T> {
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

error[E0271]: type mismatch resolving `for<'r, 's> <[closure@src/main.rs:7:18: 7:32] as std::ops::FnOnce<(&'r i32, &'s i32)>>::Output == i32`
 --> src/main.rs:8:15
  |
8 |     let foo = Foo {
  |               ^^^ expected bound lifetime parameter, found concrete lifetime
  |
note: required by `Foo`
 --> src/main.rs:1:1
  |
1 | struct Foo<T, F: Fn(&T, &T) -> T> {
  | ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

为什么错误消息说一种类型比另一种类型更通用，同时又说它们是同一类型?

Why is the error message saying that one type is more general than the other, while also saying that they are the same type?

类型仅在生存期内有所不同.夜间消息不包含生命周期-可能是在生命周期无关紧要的情况下试图降低噪音.显然，当生命周期是两种类型之间的唯一区别时，这根本没有帮助.

The types differ only in lifetimes. The nightly message doesn't include lifetimes — perhaps in an attempt to reduce noise in cases where the lifetimes are not relevant. Obviously this is not at all helpful when lifetimes are the only difference between the types.

考虑向Rust团队报告错误.

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