TypeScript 如何根据我提供的参数使我的函数通用？

Question

我想编写接受通用参数的通用函数，例如 Data<string, number>

代码如下：

interface Data<T,R> {
    a:T;
    c:R;
}

function foo(
  data: Data<string, number>
){
 return test(data)
}

//this func should be generic
function test<?, T extends Data<?, ?>>(
  data: T,
): T {
  return data
}

有什么想法吗？

Answer 1

给你：

interface Data<T,R> {
    a:T;
    c:R;
}

function foo(
  data: Data<string, number>
){
 return test(data)
}

//this func should be generic
function test<T,R>(
  data: Data<T,R>,
): Data<T,R> {
  return data
}

https://www.typescriptlang.org/play?ts=4.5.4#code/FASwdgLgpgTgZgQwMZQAQBEEQQHgCoA0ASgHyoDewq1qCAXHgNxU1J1HMC+wPcArmCQQQAezCo4IkQAoWAEyz0MinAGcIMcAHMCqMHwC2AI1glgASkqoYUCHxjjo66Quzng3HsH6DhY1E4Q+LqkstSuSpjYwaQEFnTK0YSkFCxIYqoiADZQAHRZIlouiu7UNnYOqBE8nt5SYRT0AOSBTbpsAEzc5kA

TypeScript 如何根据我提供的参数使我的函数通用？

TypeScript how can I make my function generic based on the argument I provide?

generics

types

typescript