“预期的 fn 项，发现不同的 fn 项"；使用函数指针时 [英] "Expected fn item, found a different fn item" when working with function pointers

问题描述

我有以下代码(Playground):

// Two dummy functions, both with the signature `fn(u32) -> bool`
fn foo(x: u32) -> bool {
    x % 2 == 0
}
fn bar(x: u32) -> bool {
    x == 27
}


fn call_both<F>(a: F, b: F)
where
    F: Fn(u32) -> bool,
{
    a(5);
    b(5);
}

fn main() {
    call_both(foo, bar);  // <-- error
}

对我来说，这似乎应该编译为 foo 和 bar 具有相同的签名:fn(u32) ->布尔.但是，我收到以下错误:

To me, it seems like this should compile as foo and bar have the same signature: fn(u32) -> bool. Yet, I get the following error:

error[E0308]: mismatched types
  --> src/main.rs:20:20
   |
20 |     call_both(foo, bar);
   |                    ^^^ expected fn item, found a different fn item
   |
   = note: expected type `fn(u32) -> bool {foo}`
              found type `fn(u32) -> bool {bar}`

使用此代码可以触发相同的错误:

The same error can be triggered with this code:

let mut x = foo;
x = bar;  // <-- error

我还尝试再次将 bar 强制转换为函数指针类型:

I also tried to cast bar to the function pointer type again:

let mut x = foo;
x = bar as fn(u32) -> bool;  // <-- error

这导致了一个稍微不同的错误:

This resulted in a slightly different error:

error[E0308]: mismatched types
  --> src/main.rs:20:9
   |
20 |     x = bar as fn(u32) -> bool;
   |         ^^^^^^^^^^^^^^^^^^^^^^ expected fn item, found fn pointer
   |
   = note: expected type `fn(u32) -> bool {foo}`
              found type `fn(u32) -> bool`

我完全不明白这些错误.什么是 fn 项与 fn 指针，为什么 foo 和 bar 是不同的 fn 项?

I don't understand these errors at all. What are fn items vs. fn pointers and why are foo and bar different fn items?

推荐答案

自 Rust PR #19891 被合并.但是，您可以使用 as 运算符将函数转换为相应的函数指针类型.

Each named function has a distinct type since Rust PR #19891 was merged. You can, however, cast the functions to the corresponding function pointer type with the as operator.

call_both(foo as fn(u32) -> bool, bar as fn(u32) -> bool);

只转换第一个函数也是有效的:转换将在第二个函数中被推断出来，因为两个函数必须具有相同的类型.

It's also valid to cast only the first of the functions: the cast will be inferred on the second, because both functions must have the same type.

call_both(foo as fn(u32) -> bool, bar);

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