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Episode 11 : setTimeout Closures Interview Question

Time, tide and Javascript wait for none.

  • function x() {
        var i = 1;
        setTimeout(function() {
            console.log(i);
        }, 3000);
        console.log("Namaste Javascript");
    }
    x();
    // Output:
    // Namaste Javascript
    // 1 // after waiting 3 seconds
    • 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 function inside setTimeout forms a closure (remembers reference to i). So wherever function 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.
  • Q: Print 1 after 1 sec, 2 after 2 sec till 5 : Tricky interview question

    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("Namaste Javascript");
    }
    x();
    // Output:
    // Namaste Javascript
    // 6
    // 6
    // 6
    // 6
    // 6
    • Reason?

      • This happens because of closures. When setTimeout stores the function somewhere and attaches timer to it, the function remembers its reference to i, not value of i. All 5 copies of function 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 avoid this, we can use let instead of var as let has Block scope. For each iteration, the i is a new variable altogether(new copy of i). Everytime setTimeout is run, the inside function forms closure with new variable i

    • But what if interviewer ask us to implement using var?

      function x() {
          for(var i = 1; i<=5; i  ){
          function close(i) {
              setTimeout(function() {
              console.log(i);
              }, i*1000);
              // put the setT function 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("Namaste Javascript");
      }
      x();

Watch Live On Youtube below:

setTimeout   Closures Interview Question in JS Youtube Link