JS Assignment

JavaScript Assignment: Standard, Compound, and Logical Operators

JavaScript assignment operators store values in variables and update existing bindings concisely. Beyond the basic = operator, JavaScript includes math shortcuts like += and modern ES2021 logical assignment operators like ??=.

Compound and Logical Assignment Reference

Modern Logical Assignment in Action

Logical assignment operators short-circuit, meaning they only assign when needed. This avoids overwriting valid 0 or false settings and skips unnecessary property writes.

const config = { retries: 0, label: "" };

// ??= keeps 0 because 0 is neither null nor undefined
config.retries ??= 3;
config.timeoutMs ??= 2500;

// ||= replaces empty string "" because "" is falsy
config.label ||= "Default Service";

let score = 50;
score += 15; // 65

console.log({ config, score });

Chaining Assignments Carefully

Although let a = b = 5; works in loose mode, it secretly turns b into an accidental global variable if b was not declared earlier. Declare each variable on its own line with let or const instead.

Compound assignment also works with strings (+=), remainders (%=), and exponentiation (**=). Keep each update focused on one clear variable so debugging stays simple.

Common Mistakes to Avoid

• Confusing = with === inside if: Writing if (isReady = true) assigns true instead of checking it, so the branch always runs.
• Using += on uninitialized numbers: Starting with let sum; and running sum += 5 yields NaN because undefined + 5 is NaN. Always initialize numbers to 0.
• Using ||= for numeric options: Writing options.limit ||= 10 overwrites 0 with 10; use ??= instead.

Frequently Asked Questions

What does ??= do differently from ||=?

??= assigns the right-hand value only when the left side is null or undefined. ||= assigns whenever the left side is any falsy value, including 0, false, or "".

Can I use += with const variables?

No, += reassigns the variable binding, so using += on a const variable throws a TypeError. Use let for variables that accumulate values.

Does &&= skip calling the right-hand side if the left side is falsy?

Yes, x &&= compute() short-circuits and never calls compute() when x is falsy.