Question

我正在尝试编写一个返回值的函数，或者如果该值为null或未定义，则应返回一个默认值。

Type 'A' is not assignable to type 'B | NonNullable<A>'.
  Type 'A' is not assignable to type 'NonNullable<A>'.

我收到错误

     import java.util.Scanner;

    public class Practise2{
    public static void main(String args[]){
        Scanner scan = new Scanner(System.in);

        int n = scan.nextInt();
        int m = scan.nextInt();
        int ArrForN[] = new int[n];
        int ArrForM[] = new int[m];
        int numb = 0 , pass = 1, store = 0 , val = 1 ;

        for(int i = 0; i < n; i++)
            ArrForN[i] = scan.nextInt();

        for(int j = 0; j < m ; m++)
            ArrForM[j] = j+1; 
        int y = 0;

        for(int x = 0 ; x < n ; x++){
              if(x < val*m) {    
                if(ArrForN[x] == 0){
                    ArrForN[x] = ArrForM[y];
                    numb++ ;
                    pass = pass* numb;
                    store = pass;
                    System.out.println(store);
                    y++ ;
                }
                else if(ArrForN[x] != 0)
                    y++ ;
              }
             else if(x == val*m){
               if(ArrForN[x] == 0){ 
                 System.out.println(store);  
                 val++ ;
                 numb = 0; 
                 pass = 0;
                 y = 0;    
                 ArrForN[x] = ArrForM[y];
                 numb++ ;
                 pass = pass* numb;
                 store = pass;
                 System.out.println(store); 
                 y++ ;
               }  
               else if(ArrForN[x] != 0)  
                   y++ ;

            }
        } 

    }
 }

NonNullable应该为A，不能为null或未定义，这就是我在if中检查的内容吗？

Here is the code in ts playground.

Answer 1

条件类型（其中NonNullable是）通常不能在泛型函数中提供良好的实现。您可以使用类型断言来使其生效（return input as any）;

一种更安全的方法可能是稍微改变一下类型：

function test<A, B>(input: A | undefined | null, fallbackValue: B): A | B {
    if (input == null || input == undefined) {
        return fallbackValue;
    } else {
        return input;
    }
}

declare var s: string | null;
let a = test(s, "") // string

Typescript函数返回的值没有null或undefined

1 个答案: