[RxJS] defer() lazy evaluation

For example, we have following code:

复制代码
import { of, defer} from 'rxjs'; 

class Foo {
  private num = 123
  onum = of(this.num);

  updateNum(val) {
    this.num = val;
  }
}

const f = new Foo();
f.onum.subscribe(x => console.log(x)) // 123
复制代码

 

If I want to update the 'num' before subscribe:

复制代码
import { of, defer} from 'rxjs'; 

class Foo {
  private num = 123
  onum = of(this.num);

  updateNum(val) {
    this.num = val;
  }
}

const f = new Foo();
f.updateNum(321)
f.onum.subscribe(x => console.log(x)) // 123
复制代码

The final value is still '123'.

 

This is because 'of()' remember the value during the construction time. So it is always 123.

 

How to solve the problem? Using defer.

复制代码
import { of, defer} from 'rxjs'; 


class Foo {
  private num = 123
  onum = defer(() => of(this.num));

  updateNum(val) {
    this.num = val;
  }
}

const f = new Foo();
f.updateNum(321)
f.onum.subscribe(x => console.log(x)) // 321
复制代码

 

'defer' lazy evaluate the value during the time we call subscribe

posted @   Zhentiw  阅读(336)  评论(0编辑  收藏  举报
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