Question

我正在尝试编写阶乘函数，但是出现了可怕的语法错误。我将问题简化为几行代码。我曾尝试将值分配更改为let value =，但是最后我遇到了无限循环。

extern crate num; // 0.2.0

use num::{bigint::BigUint, One};

fn decrease(mut value: &BigUint) {
    while value != &BigUint::one() {
        value = &(value - BigUint::one());
        println!("new value {}", value);
    }
}

error[E0597]: borrowed value does not live long enough
 --> src/lib.rs:7:18
  |
7 |         value = &(value - BigUint::one());
  |                  ^^^^^^^^^^^^^^^^^^^^^^^^- temporary value only lives until here
  |                  |
  |                  temporary value does not live long enough
  |
note: borrowed value must be valid for the anonymous lifetime #1 defined on the function body at 5:1...
 --> src/lib.rs:5:1
  |
5 | / fn decrease(mut value: &BigUint) {
6 | |     while value != &BigUint::one() {
7 | |         value = &(value - BigUint::one());
8 | |         println!("new value {}", value);
9 | |     }
10| | }
  | |_^
  = note: consider using a `let` binding to increase its lifetime

Answer 1

如果要更改value的值...，必须将其视为&mut。认为mut &毫无意义。在您可以使用-=之后。

extern crate num; // 0.2.0

use num::{BigUint, One};

fn decrease(value: &mut BigUint) {
    while value != &BigUint::one() {
        *value -= BigUint::one();
        println!("new value {}", value);
    }
}

fn main() {
    decrease(&mut BigUint::new(vec!(42)));
}

由于临时值的寿命不足，因此无法在循环中递减＆BigUint

1 个答案: