[Solved] of('A', 'B').pipe( concatMap(letter => interval(20).pipe(take(3), map(i => letter + i))) ).subscribe(console.log);What will be the order of emitted values? | Angular

Angular - RxJS Operators

Consider the following Angular RxJS code:

import { of, interval } from 'rxjs';
import { concatMap, take, map } from 'rxjs/operators';

of('A', 'B').pipe(
  concatMap(letter => interval(20).pipe(take(3), map(i => letter + i)))
).subscribe(console.log);

What will be the order of emitted values?

AB0, A0, B1, A1, B2, A2

BA0, B0, A1, B1, A2, B2

CB0, B1, B2, A0, A1, A2

DA0, A1, A2, B0, B1, B2

Step-by-Step Solution

Solution:

Step 1: Understand concatMap
concatMap queues inner observables and subscribes to them one after another.
Step 2: Analyze emissions
First, 'A' triggers interval emitting A0, A1, A2 sequentially; only after completion does 'B' start emitting B0, B1, B2.
Final Answer:
A0, A1, A2, B0, B1, B2 -> Option D
Quick Check:
concatMap preserves order by waiting for each inner observable [OK]

Quick Trick: concatMap processes inner observables sequentially [OK]

Common Mistakes:

MISTAKES

Assuming emissions from inner observables interleave
Confusing concatMap with mergeMap behavior

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More Angular Quizzes

Consider the following Angular RxJS code:

Step 1: Understand concatMap

Step 2: Analyze emissions

Final Answer:

Quick Check:

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