CODEWITHSUNDEEP
javascriptecmascript-6let
Eggs
Eggs

Reputation: 345

ES6: Multiple let assignment inside for-loop in 'strict mode'

I am trying out 'strict mode' and using keyword let inside of a for-loop like this: 

  (function() {

    "use strict";

    var numbers = [1,2,3,4,5];
    var value;


    for (let i = j = 0; i < numbers.length; i++) {
      j += numbers[i];
      value = j;
    }
      console.log ("\tThe reduced value is: ", value);    
  })();

This produces an Uncaught ReferenceError: j is not defined, but by removing 'use strict' the code runs fine.

The code is reduced subset of my jsfiddle

Upvotes: 1

Views: 1226

Answers (2)

keithlee96
keithlee96

Reputation: 1944

The problem is as you would expect, the initialiser statement in your for loop:

let i = j = 0;

This statement is identical to:

let i = (j=0);

which is the same as:

j = 0; // would throw : Uncaught ReferenceError: j is not defined
let i = j;

If you want to fix the strict-mode error, you need to explicitly declare j as a variable with let j = 0;

It works in sloppy mode due to hoisting, which will create a variable j in the global scope. (https://www.w3schools.com/js/js_hoisting.asp)

Edit: Bergi is right, I mixed up the concepts, it becomes available in the global scope due to the the scope chain, not hoisting. If a variable is undeclared in the local scope, it looks at the outer lexical scope for the variable. If it's still not defined there, it keeps going up the scope chain until it hits the global scope. If it's still not defined there, then it automatically gets defined in the global scope and that is the variable you reference every time you use j.

Upvotes: 3

smnbbrv
smnbbrv

Reputation: 24541

The construction 

let i = j = 0

has actually nothing to do with the keyword let. Yet

Let's look at what it does under the hood:

let i // declares i
j = 0 // assigns a variable j to 0 that is not declared
      // If it does not exist it creates the variable on the `window` object.
i = j // in fact i = window.j

To make it right use 

let i = 0, j = 0

Upvotes: 3

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