function x() {
  var i = 1;
  setTimeout(function() {
  console.log(i);
  }, 3000);
  console.log("This is Hari");
 }
 x();
 

This is Hari

1 //after waiting 3 seconds (3000ms)

We expect JS to wait 3 sec, print 1 and then go down and print the string. But JS prints string immediately, waits 3 sec and then prints 1.

• The fun inside setTimeout forms a closure (remembers reference to i). So wherever fun goes it carries this ref along with it.
• setTimeout takes this callback function & attaches timer of 3000ms and stores it. Goes to next line without waiting and prints string.
• After 3000ms runs out, JS takes function, puts it into call stack and runs it.

We assume this has a simple approach as below

function x() {
 for(var i = 1; i<=5; i++){
   setTimeout(function() {
     console.log(i);
    }, i*1000);
   }
   console.log("This is Hari");
 }
 x();
 

This is Hari

6

6

6

6

6

• This happens because of closures. When setTimeout stores the function somewhere and attaches timer to it, the fun remembers its reference to i, not value of i
• All 5 copies of fun point to same reference of i.
• JS stores these 5 functions, prints string and then comes back to the functions. By then the timer has run fully. And due to looping, the i value became 6. And when the callback fun runs the variable i = 6. So same 6 is printed in each log
• To stop this from happening, use let instead of var as let has black scope. For each iteration, the i is a new variable altogether(new copy of i).
• Everytime setTimeout is run, the inside fun forms closure with new variable i

function x() {
 for(var i = 1; i<=5; i++){
   function close(i) {
    setTimeout(function() {
     console.log(i);
    }, i*1000);
    // put the setT fun inside new function close()
   }
   close(i); // everytime you call close(i) it creates new copy of i. Only this time, it is with var itself!
  }
   console.log("This is Hari");
 }
 x();