What is the best way to utilize a union of interfaces in TypeScript when working with instances?

Review this TypeScript snippet:

class A {public name: string = 'A';}
interface B {num: number;}

class Foo {
    get mixed(): Array<A | B> {
        const result = [];
        for (var i = 0; i < 100; i ++) {
            result.push(Math.random() > 0.5 ? new A() : {num: 42});
        }
        return result;
    }

    print(): string {
        return this.mixed.map(item => {
            if (item instanceof A) {
                return item.name;
            }

            return item.num;    
        }).join('\n');
    }
}

In the print method, different values need to be returned based on the type of item. While it's easy to use instanceof for type A, the use of B as an interface makes it challenging.

TypeScript may not recognize that item is definitely of type

B</code when accessing <code>item.num
, leading to unnecessary complaints.

Answer №1

If you want to nudge the TypeScript compiler in the right direction, you can utilize the following code snippet:

return (<B>item).num;

This will provide a helpful hint to the TypeScript compiler.

Answer №2

Another effective approach is to utilize user-defined type guard functions

This technique involves creating custom type guard functions like the following example:

class A ...
interface B ...
function isB(item: any): item is B {
    return item && 'num' in item;
}

class Foo {...
   print(): string {
       return this.mixed.map(item => {
          if (item instanceof A) {
            return item.name;
          } else if(isB(item)){
            return item.num;
          }  
  }).join('\n');
}..

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