我有一个函数可以与枚举一起使用来应用二进制函数。这是给口译员的:
use std::ops::*;
#[derive(Debug, Clone, PartialEq, PartialOrd)]
pub enum Scalar {
I64(i64),
I32(i32),
//many many others
}
pub trait TMath: Add + Mul + Sized {} //mark numerical types
impl<T: Add + Mul> TMath for T {}
fn add<T: TMath>(x: T, y: T) -> <T as Add>::Output {
x + y
}
pub type NatBinExpr<T: TMath> = Fn(&T, &T) -> T;
我想做的事:
let result = bin_op(add, &Scalar::I32(1), &Scalar::I32(2));
同时也让它适用于任意二元函数:
let result = bin_op(Scalar::concat, &Scalar::I32(1), &Scalar::I32(2));
但是,我还没有找到一种方法来传递闭包而不使 bin_op
通用:
fn bin_op(apply: &NatBinExpr???, x: &Scalar, y: &Scalar) -> Scalar {
match (x, y) {
(Scalar::I64(a), Scalar::I64(b)) => Scalar::I64(apply(a, b)),
(Scalar::I32(a), Scalar::I32(b)) => Scalar::I32(apply(a, b)),
}
}
使 bin_op
通用化是不对的; bin_op
对Scalar
进行操作,但内部操作是通用的。
最佳答案
基本上有两种不同的方式来讨论函数类型:
- 指针:
fn(A, B) -> C
, - 特征:
Fn(A, B) -> C
,FnMut(A, B) -> C
,FnOnce(A, B) -> C
.
无论哪种情况,它们都以参数和结果类型为特征。
那么,apply
的参数和结果类型是什么?
视情况而定。
从您的示例中,我们可以看到 [i64, i32, ... 中的
。T
是 FnOnce(T, T) -> T
]
这不是一个类型,而是许多类型。因此它需要的不是单一的功能,而是多种功能;或者可能是多次实现 FnOnce
的函数对象。
函数对象路由仅在夜间可用,并且需要大量样板文件(宏对此有帮助):
#![feature(fn_traits)]
#![feature(unboxed_closures)]
use std::ops::*;
#[derive(Debug, Clone, PartialEq, PartialOrd)]
pub enum Scalar {
I64(i64),
I32(i32),
//many many others
}
pub trait TMath: Add + Mul + Sized {} //mark numerical types
impl<T: Add + Mul> TMath for T {}
struct Adder;
impl FnOnce<(i64, i64)> for Adder {
type Output = i64;
extern "rust-call" fn call_once(self, args: (i64, i64)) -> i64 {
args.0 + args.1
}
}
impl FnMut<(i64, i64)> for Adder {
extern "rust-call" fn call_mut(&mut self, args: (i64, i64)) -> i64 {
args.0 + args.1
}
}
impl Fn<(i64, i64)> for Adder {
extern "rust-call" fn call(&self, args: (i64, i64)) -> i64 {
args.0 + args.1
}
}
impl FnOnce<(i32, i32)> for Adder {
type Output = i32;
extern "rust-call" fn call_once(self, args: (i32, i32)) -> i32 {
args.0 + args.1
}
}
impl FnMut<(i32, i32)> for Adder {
extern "rust-call" fn call_mut(&mut self, args: (i32, i32)) -> i32 {
args.0 + args.1
}
}
impl Fn<(i32, i32)> for Adder {
extern "rust-call" fn call(&self, args: (i32, i32)) -> i32 {
args.0 + args.1
}
}
fn bin_op<F>(apply: &F, x: Scalar, y: Scalar) -> Scalar
where
F: Fn(i64, i64) -> i64,
F: Fn(i32, i32) -> i32,
{
match (x, y) {
(Scalar::I64(a), Scalar::I64(b))
=> Scalar::I64((apply as &Fn(i64, i64) -> i64)(a, b)),
(Scalar::I32(a), Scalar::I32(b))
=> Scalar::I32((apply as &Fn(i32, i32) -> i32)(a, b)),
_ => unreachable!(),
}
}
fn main() {
let result = bin_op(&Adder, Scalar::I32(1), Scalar::I32(2));
println!("{:?}", result);
}
关于generics - 如何在不使函数泛型的情况下将具有泛型的闭包传递给函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52539771/