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!== (strict inequality) operator evaluates whether two operands are not identical. It returns true if the operands differ in either data type or value, and false if they share the exact same type and value, with the notable exception of NaN. Unlike the loose inequality operator (!=), !== never performs implicit type coercion during evaluation.
Evaluation Mechanics
At runtime, the operator follows strict equality resolution rules:- Type Comparison: If the runtime types of
operand1andoperand2differ, it immediately evaluates totrue. - Value Comparison: If the types are identical, it compares the underlying values. If the values differ, it evaluates to
true. - The
NaNException: In JavaScript and TypeScript,NaN(Not-a-Number) is of typenumberbut is strictly unequal to itself. Therefore,NaN !== NaNevaluates totrue. To accurately check if a value isNaN, you must useNumber.isNaN()rather than the!==operator. - Reference Types: For objects, arrays, and functions,
!==evaluates memory identity (reference equality) rather than structural equivalence. Two distinct objects with identical properties will evaluate totruebecause they occupy different memory addresses.
TypeScript Compiler Behavior
While the runtime behavior is identical to JavaScript, TypeScript applies static type analysis to the!== operator at compile time.
If the TypeScript compiler determines that the types of the two operands have no intersection (i.e., they are mutually exclusive or non-overlapping), it will throw a ts(2367) compiler error. TypeScript flags this because comparing two completely unrelated types with !== will always evaluate to true, indicating a logical flaw in the code.
any or unknown.
Reference Evaluation Example
When evaluating reference types, the compiler allows the comparison as long as the types overlap, but the runtime evaluation strictly checks the memory pointer.Master TypeScript with Deep Grasping Methodology!Learn More





