What is the output of foo.call(obj, 5) where foo returns this.x + y and obj.x is 10?
15. call sets this to obj (so this.x is 10) and passes 5 as y. 10 + 5 = 15.
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More FAQs in call, apply, and bind Interview Questions
No. The first bind wins. bound.bind(newObj) does not change this. The bound function keeps the original this from the first bind call.
Array.prototype.slice.call(arguments). This borrows the slice method from Array.prototype and applies it to the array-like arguments object.
Nothing. The arrow function's this is lexical (inherited from the enclosing scope). bind, call, and apply cannot change an arrow's this.
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