Does destructuring break a closure in JavaScript?
No. Destructuring methods from a closure-based object (const { increment, getCount } = c) still close over the same private variables. Both increment and getCount access the same count. The closure is not broken.
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More FAQs in Closures Deep Dive: Interview Questions in JavaScript
Create a cache object in the outer function. Return a function that serializes the arguments as a key, checks the cache, and either returns the cached result or calls the original function and stores the result. The cache is private to the closure.
Close over a timer variable. Each call clears the previous timer and sets a new one. The function only runs after the user stops calling for the specified delay. The timer persists across calls via the closure.
Each nested function closes over its parent's variables. So makeAdder(5)(3)(2) works because the innermost function can access x (from the outermost), y (from the middle), and z (its own parameter). The scope chain connects all of them.
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