The answer
Because arrays are passed by reference. Mutating the parameter mutates the same object the caller holds. Copy the array ([...list]) before mutating if you want to preserve the original.
Common Function Bugs in JavaScript and How to Fix Them
Why does a function that mutates its array parameter also change the caller's array?
More FAQs in Common Function Bugs in JavaScript and How to Fix Them
Because the method is detached from obj when passed as a reference. setTimeout calls it with this as the global object (or undefined in strict mode). Fix it with .bind(obj), an arrow wrapper, or () => obj.method().
Use let instead of var so each iteration gets a fresh binding, or wrap the body in an IIFE that captures the current value of i as a parameter.
Because the braces are interpreted as a block body, and the object literal looks like a label statement. Wrapping the object in parentheses (n) => ({ x: 1 }) makes it an expression body that returns the object.
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