JS written test question -- promise, setTimeout, task queue comprehensive question

			setTimeout((n1) => {
				console.log(0);
			});  
			new Promise(resolve => {
				console.log(1); 
				setTimeout((n2) => {    
					resolve(); 
					var p1=new Promise((n1,n2)=>{n1(20)}) 
					p1.then((t1) => console.log(2));
					console.log(3);
					setTimeout((n3)=>{console.log(9)});
				}); 
				new Promise((n1,n2)=>{n1(20)}).then((t2) => console.log(4));
				
			}).then((t3) => {   
				console.log(5); 
				var p2=new Promise((n1,n2)=>{n1(20)}) 
				p2.then((t4) => console.log(8)); 
				setTimeout((n4) => console.log(6)); 
			});  
			console.log(7);

here setTimeout With n1~n4 As parameter marker ;then With t1~t4 As a mark

Knowledge point ：
1.then It belongs to asynchronous micro task ,setTimeout It belongs to asynchronous macro task ,
2. The queue priority of asynchronous tasks is in the same cycle : The asynchronous macro task executes first Then execute asynchronous micro tasks
3. The asynchronous macro task is to start the second thread and wait for the next cycle to start before execution , Asynchronous micro tasks are executed at the end of this cycle
Execution order ：
Script run Perform the first macro task :
1. Perform the synchronization task first
2. Add a new macro task to the queue Add a new asynchronous micro task
3. Executing asynchronous micro tasks

Analysis by graph ： The first cycle begins ： Perform the synchronization task first , So print 1,7,; Then execute asynchronous micro tasks , Print first 4, because then(t3) Not in this cycle ;
Use resolve() To pass in data , So do not execute and wait in the asynchronous micro task queue . The second cycle begins , Execute asynchronous macro task queue , Then a new synchronization task is generated , Print first 0,3; because resolve() Called , therefore then(t3) Execute the generate synchronization task Print 5 , Asynchronous tasks are also generated here , See who gets to the asynchronous task queue first and executes it first ; the reason being that setTimeout(n2) Internally invoked resolve(), therefore then(t1) and setTimeout(n3) In front of each other , Execute asynchronous micro task printing 2,8
Then execute asynchronous macro task printing 9,6