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Can Logical Intersections Evaluate to True with Conflicting Conditions?
In an intriguing interview query posed by a renowned tech corporation, the question arises: can the expression (a == 1 && a == 2 && a == 3) ever evaluate to true in JavaScript?
Initially, this may seem implausible given the seemingly contradictory nature of the conditions. However, by delving into the intricacies of JavaScript's equality operator (==), a path emerges towards answering this enigma.
JavaScript's equality operator, unlike its strictly equal counterpart (===), performs type coercion, automatically converting values to a common type for comparison. This opens the door to the manipulation of objects, particularly with customized toString (or valueOf) functions, which allow for dynamic value generation.
Consider the following example:
const a = { i: 1, toString: function () { return a.i++; } }
With this custom toString function, each time a is implicitly coerced to a string (e.g., for logging), a's internal i value increments. This provides a mechanism to satisfy all three equality conditions:
a == 1 // true (initially i = 1) a == 2 // true (after toString call, i = 2) a == 3 // true (after second toString call, i = 3)
Thus, by leveraging the dynamic behavior of objects and the implicit type conversion of ==, it becomes possible for the expression (a == 1 && a == 2 && a == 3) to evaluate to true in JavaScript.
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