Interactive JavaScript interview practice

Execution Context Interview Lab

Practice real JavaScript interview questions with the DeepJS execution visualizer. Predict the output first, then trace exactly how the result is produced.

  1. Why does the local variable log undefined?

    Declaration instantiation

    A var declaration inside a function creates a local binding before its assignment is evaluated.

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  2. What happens when a parameter and a local var share a name?

    Lexical scope

    A var declaration that names an existing parameter does not create a new binding — it reuses the parameter's binding.

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  3. Does a function read variables from its caller, or from where it was defined?

    Lexical scope

    JavaScript uses lexical (static) scoping, so a function's scope chain is fixed by where it is written, not by who calls it.

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  4. Can global code read a variable declared inside a function?

    Lexical scope

    A function-local var binding is not visible from the outer global environment.

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  5. Why is x initially a function and later a number?

    Declaration instantiation

    Declaration instantiation creates one binding, then later assignments replace the value stored in it.

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  6. Why does var escape the if block?

    Declaration instantiation

    var is scoped to a function or script, not to an if block.

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  7. When can an undeclared assignment create a global?

    Global bindings

    Sloppy script assignment to an unresolvable name can create a property on the global object.

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  8. Which assignments affect parameters, locals, and globals?

    Lexical scope

    The parameter a and local b shadow globals, while c resolves to the outer binding.

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  9. How can a function binding become a number?

    Function calls

    The function declaration initializes a, then the var initializer overwrites that same binding with 2.

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  10. Does changing a parameter change the outer primitive?

    Function calls

    The parameter receives a copy of the number value, stored in a separate function binding.

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  11. How does a closure preserve changing state?

    Closures

    The returned function keeps access to fn’s n binding after fn has returned.

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  12. Why do all three setTimeout callbacks log 3, not 0, 1, 2?

    Closures

    var creates one shared binding for the whole loop, so every callback closes over the same final value.

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  13. Why does swapping var for let fix the setTimeout loop?

    Closures

    let creates a fresh binding for i on every iteration, so each callback closes over its own private value.

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  14. Can an IIFE fix the setTimeout loop without using let?

    Closures

    Calling an immediately invoked function with i as its argument gives each iteration its own parameter binding.

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