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「重学TS 2.0 」TS 练习题第三十五题
实现一个 Add 工具类型，用于实现对数值类型对应的数值进行加法运算。具体的使用示例如下所示：
type Add<T, R> = // 你的实现代码
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
请在下面评论你的答案
type Push<T extends any[], V> = [...T, V];
type CreateTuple<
T extends number,
A extends any[] = []
> = A['length'] extends T ? A : CreateTuple<T, Push<A, ''>>
type Add<
T extends number,
R extends number
> = [...CreateTuple<T>, ...CreateTuple<R>]['length']
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
没啥难的，就是通过数值，构建对应长度的数组的常规操作
// 构建数组
type GenArr<T extends number, S extends any[] = []> = S['length'] extends T ? S : GenArr<T, [...S, 1]>
type Add<T extends number, R extends number> = [...GenArr<T>, ...GenArr<R>]['length']
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
type Add<
N extends number,
M extends number,
S extends any[] = [],
Q1 extends any[] = [],
Q2 extends any[] = [],
L1 = Q1["length"],
L2 = Q2["length"],
L = S["length"],
> = L1 extends N
? L2 extends M
? L
: Add<N, M, [...S, 1], Q1, [...Q2, 1]>
: Add<N, M, [...S, 1], [...Q1, 1], Q2>;
// 测试用例
type S0 = Add<100, 200>; // 300
使用之前所写的类型
type Add<T extends number, R extends number> = Len<
[...Repeat<" ", T>, ...Repeat<" ", R>]
>;
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20>; // 28
type A2 = Add<10, 30>; // 40
type Add<T, R, A1 extends any[] = [], A2 extends any[] = [], AT extends any[] = []> = A1["length"] extends T
? A2["length"] extends R
? AT["length"]
: Add<T, R, A1, [...A2, ""], [...AT, ""]>
: Add<T, R, [...A1, ""], A2, [...AT, ""]>
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
思路: 依然是构造数组长度
export default {}
// 实现一个 Add 工具类型，用于实现对数值类型对应的数值进行加法运算。具体的使用示例如下所示：
type Add<T, R, TArr extends any[] = [], RArr extends any[] = []> = ( TArr['length'] extends T ? RArr['length'] extends R ? [...TArr, ...RArr]['length'] : Add<T, R, TArr, [...RArr, 1]> : Add<T, R, [...TArr, 1], RArr> )
type A0 = Add<5, 5>; // 10 type A1 = Add<8, 18> // 28 type A2 = Add<10, 15>; // 40
type Add<T, R, Arr1 extends any[] = [], Arr2 extends any[] = []> = Arr1['length'] extends T ? Arr2['length'] extends R ? [...Arr1, ...Arr2]['length'] : Add<T, R, Arr1, [...Arr2, '']> : Add<T, R, [...Arr1, ''], Arr2>
type A0 = Add<5, 5>; // 10 type A1 = Add<8, 20> // 28 type A2 = Add<10, 30>; // 40
type Add<T, R, Arr1 extends any[] = [], Arr2 extends any[] = []> = Arr1['length'] extends T ? Arr2['length'] extends R ? [...Arr1, ...Arr2]['length'] : Add<T, R, Arr1, [...Arr2, '']> : Add<T, R, [...Arr1, ''], Arr2>
type A0 = Add<5, 5>; // 10 type A1 = Add<8, 20> // 28 type A2 = Add<10, 30>; // 40
type BuildArray<
T extends number,
C extends number[] = []
> = C["length"] extends T ? C : BuildArray<T, [...C, number]>;
type Add<T extends number, R extends number> = [
...BuildArray<T>,
...BuildArray<R>
]["length"];
// 类似前面两题, 需要构建数值对应的数组, 然后返回该数组的长度即可
type NumToArr<T, L extends number = 0, R extends any[] = []> = T extends L ? R : NumToArr<T, R['length'], [...R, '']>
type Add<T, R> = [...NumToArr<T>, ...NumToArr<R>]['length']
type loop<T, U extends any[] = []> = U['length'] extends T ? U : loop<T,[...U, 1]>
type Add<T extends number, R extends number> = [...loop<T>, ...loop<R>]['length']
// 你的实现代码
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
type BuildArr<T extends number, R extends any[] = []> = R['length'] extends T ? R: BuildArr<T, [...R, 1]>;
type Add<T extends number, R extends number> = [...BuildArr<T>, ...BuildArr<R>]['length'] // 你的实现代码
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
type Add<T, R, A extends any[] = [], B extends any[] = [],> = A['length'] extends T ?
B['length'] extends R ? [...A, ...B]['length'] : Add<T, R, A, [...B, '']>
: Add<T, R, [...A, ''], B>
type A0 = Add<5, 5>; // 10
type A1 = Add<8, 20> // 28
type A2 = Add<10, 30>; // 40
