如果在函数内创建引用，如何将泛型类型与需要使用生命周期参数的特征绑定在一起？

Question

I want to implement a generic fibonacci function that works with any type implementing Zero , One , and AddAssign . I first implemented a version that works fine, but is specialized for num::BigUint ( see on play.rust-lang.org ). I than came up with the following generic implementation ( see on play.rust-lang.org ):

extern crate num;

use num::{One, Zero};
use std::mem::swap;
use std::ops::AddAssign;

fn fib<T: Zero + One + AddAssign<&T>>(n: usize) -> T {
    let mut f0 = Zero::zero();
    let mut f1 = One::one();
    for _ in 0..n {
        f0 += &f1;
        swap(&mut f0, &mut f1);
    }
    f0
}

This doesn't compile:

error[E0106]: missing lifetime specifier
 --> src/main.rs:7:34
  |
7 | fn fib<T: Zero + One + AddAssign<&T>>(n: usize) -> T {
  |                                  ^ expected lifetime parameter

Rust wants me to add a lifetime parameter to AddAssign<&T> but I don't know how to express the lifetime of f1 .

Answer 1

You need to use Higher Rank Trait Bounds . This one means basically "For any lifetime 'a , T satisfies the AddAssign<&'a T> trait":

fn fib<T>(n: usize) -> T
where
    for<'a> T: Zero + One + AddAssign<&'a T>,

I also had to change the way fib is called because the compiler couldn't figure out the return type, which could be literally any type that implements those traits. Declaring x 's type gives sufficient context to the compiler so that it knows what you want.

fn main() {
    let x: num::BigUint = fib(10);
    // let x = fib::<BigUint>(10); // Also works
    println!("fib(10) = {}", x);
}

playground