How does the call pass its argument to the inner function?

Question

The argument 10 gets passed to the anonymous inner function, bypassing the outer function. What is the principle here?

function aFunc() {
  let firstNum = 2;
  return (secondNum) => secondNum * firstNum;
}
let aVar = aFunc();
console.log(aVar(10));

Answer 1

Because aVar is (secondNum) => secondNum * firstNum where firstNum and the closure has firstNum = 2.

function aFunc() {
  let firstNum = 2;
  return (secondNum) => secondNum * firstNum;
}
let aVar = aFunc();

Calling aFunc() runs the function which creates a closure around firstNum and returning the arrow function (secondNum) => secondNum * firstNum.

The subsequent call to aVar(10) calls this arrow function and the result of adding firstNum (2) and secondNum (10) is calculated and returned.

console.log(aVar(10));