Rust 无法实现整型类型的泛型FN

我对C++很有经验，并开始try 使用Ruust.

在try 实现一些简单的泛型函数时，我遇到了以下问题:

use std::ops::BitAnd;
use std::cmp::Eq;

fn is_odd_i32(x: u32) -> bool {
    if x & 1_u32 == 1_u32 { true } else { false }
}

fn is_odd<T: BitAnd + Eq>(x: &T) -> bool {
    if (*x & (1 as T)) == (1 as T) { true } else { false }
}

fn main() {
    println!("is_odd -> '{}'", is_odd(&23_u64));
    println!("is_odd -> '{}'", is_odd(&23_u32));
}

问题似乎是将按位AND结果与0或1进行比较.我知道要实现这一点，1(或0)必须可以转换为类型T，但不知道如何实现.我也试了T::try_from(1_u8).ok().unwrap()次，但这也不起作用.

我不知道该怎么解决这个问题.

我收到的错误是:

error[E0369]: binary operation `==` cannot be applied to type `<T as BitAnd>::Output`
  --> src/main.rs:27:24
   |
27 |     if (*x & (1 as T)) == (1 as T) { true } else { false }
   |        --------------- ^^ -------- T
   |        |
   |        <T as BitAnd>::Output
   |
help: consider further restricting the associated type
   |
26 | fn is_odd<T: BitAnd + Eq>(x: &T) -> bool where <T as BitAnd>::Output: PartialEq<T> {
   |                                          +++++++++++++++++++++++++++++++++++++++++

error[E0605]: non-primitive cast: `{integer}` as `T`
  --> src/main.rs:27:14
   |
27 |     if (*x & (1 as T)) == (1 as T) { true } else { false }
   |              ^^^^^^^^ an `as` expression can only be used to convert between primitive types or to coerce to a specific trait object

error[E0605]: non-primitive cast: `{integer}` as `T`
  --> src/main.rs:27:27
   |
27 |     if (*x & (1 as T)) == (1 as T) { true } else { false }
   |                           ^^^^^^^^ an `as` expression can only be used to convert between primitive types or to coerce to a specific trait object

顺便说一句，我只是在玩弄特征和泛型，这不是测试整数是否奇数的最好方法.