推广添加无符号和有符号整数类型

Question

我想有一个允许添加的防锈功能u32(u64,u128)类型的i32(i64,i128)类型,而检查溢出.

我的实施:

/// Add u32 to i32. In case of an overflow, return None.
fn checked_add_i32_u32(a: i32, b: u32) -> Option<i32> {
    let b_half = (b / 2) as i32;
    let b_rem = (b % 2) as i32;

    Some(a.checked_add(b_half)?.checked_add(b_half)?
        .checked_add(b_rem)?)
}

/// Add u64 to i64. In case of an overflow, return None.
fn checked_add_i64_u64(a: i64, b: u64) -> Option<i64> {
    let b_half = (b / 2) as i64;
    let b_rem = (b % 2) as i64;

    Some(a.checked_add(b_half)?.checked_add(b_half)?
        .checked_add(b_rem)?)
}

我还有另外一个类似的u128和i128.我觉得我在重复自己.我对这些功能的测试看起来非常相似.

有没有办法可以重构我的代码并只改为一个函数？我不确定如何概括u32和i32(或u64和i64,u128和i128)之间的关系.

Answer 1

你可以使用宏:

trait CustomAdd: Copy {
    type Unsigned;

    fn my_checked_add(self, b: Self::Unsigned) -> Option<Self>;
}

macro_rules! impl_custom_add {
    ( $i:ty, $u:ty ) => {
        impl CustomAdd for $i {
            type Unsigned = $u;

            fn my_checked_add(self, b: $u) -> Option<$i> {
                let b_half = (b / 2) as $i;
                let b_rem = (b % 2) as $i;

                Some(self.checked_add(b_half)?.checked_add(b_half)?
                    .checked_add(b_rem)?)
            }
        }
    }
}

impl_custom_add!(i32, u32);
impl_custom_add!(i64, u64);
// etc.

#[test]
fn tests() {
    assert_eq!(123.my_checked_add(10_u32), Some(133));
}