JavaScript Array.find() Returns Reference to Object

JavaScript objects don’t behave the way many people expect. ✔ The output of the code in the picture will be: [  { id: 1, name: "Updated John" },  { id: 2, name: "Jane" } ] This surprises many people, but it is completely expected behavior in JavaScript. 🤔Why this happens? → Arrays in JavaScript store references to objects → Array.find() returns the actual object, not a copy → Objects are reference types, not value types So after this line: const foundUser = users.find(u => u.id === 1); 👉 Both of these point to the same object in memory: users[0] ────┐        ├──► { id: 1, name: "John" } foundUser ───┘ 👉 When you do: foundUser.name = "Updated John"; You are mutating that shared object. Since the array holds a reference to the same object, the array reflects the change. 💡 A safer approach is to update immutably (create a new array and a new object): const updatedUsers = []; const updatedUsers = users.map(user =>  user.id === 1 ? { ...user, name: "Updated John" } : user ); ▶ But remember: { ...user } is a shallow copy. If user contains nested objects, those nested references are still shared. In that case, you must copy the nested structure you modify, or use a deep clone. ▶ There is another option which is called structuredClone(). This function returns a deep copy of the object you are passing as an argument. const copy = structuredClone(user); Now mutating copy won’t affect the original object. #JavaScript #WebDevelopment #Coding

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find method returns the same object reference so the log is the string "Updated John"

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