Why does let in a setTimeout loop log 0, 1, 2 in JavaScript?
Because let is block-scoped and creates a fresh binding per iteration. Each callback closes over its own copy of i, which holds the value for that iteration. So when they run, each sees its own value.
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More FAQs in The setTimeout Loop Closure Bug: var vs let
Because var is function-scoped. There is one i for the entire loop. All callbacks close over the same i. By the time they run (after the loop), i is 3. Closures close over the variable, not the value.
The variable. A closure holds a reference to the variable, not a snapshot of its value. If the variable changes before the callback runs, the callback sees the updated value. This is why var in a loop logs the final value.
By creating a new function scope per iteration and capturing the current value of i as a parameter (j). Each callback closes over its own j, which is a snapshot of i at that iteration. This was the pre-ES6 workaround.
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