How does this code work as conditional statements, why aren't there any "if" or "else if" statements?

Question

Given the following problem that I need to solve about nested logic, I saw the following possible solution online but I hadn't seen code organized like this and can't help to figure it out. It looks like a different approach I hadn't seen to "if" and "else if" statements, so I would like to understand what's going on. Thank you.

let fine = 0;
const [actual, expected] = input.split('\n').map(item => {
  const [day, month, year] = item.split(' ').map(Number);

  return {
    day,
    month,
    year
  };
});

(
  actual.year === expected.year &&
  actual.month === expected.month &&
  actual.day > expected.day
) && (fine = (actual.day - expected.day) * 15);

(
  actual.year === expected.year &&
  actual.month > expected.month
) && (fine = (actual.month - expected.month) * 500);

(actual.year > expected.year) && (fine = 10000);

console.log(fine);
}

Terrible abuse of the **short-circuiting behavior** of logical operators: `false && never_called()` — user2864740, Sep 29 '20 at 03:52
Related reading: https://stackoverflow.com/q/12554578/2864740 // https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Logical_AND // https://dev.to/akhil_001/short-circuit-evaluation-in-javascript-3o5a — user2864740, Sep 29 '20 at 03:55
Very poorly written code. It appears to be "obtuse on purpose" which is an anti-pattern. — jfriend00, Sep 29 '20 at 03:56

score 1 · Answer 1 · answered Sep 29 '20 at 03:56

JavaScript has short-circuit boolean evaluation, and the code in your question uses that extensively.

Take for example:

a() && b(); // if (a()) b();

Because both a() and b() need to evaluate to a truthy value for the entire statement to be considered true, JavaScript will not evaluate b() if a() already evaluates to a falsy value.

Similarly:

c() || d(); // if (!c()) d();

Because only one of c() or d() needs to evaluate to a truthy value for the entire statement to be considered true, JavaScript will not evaluate d() if c() already evaluates to a truthy value.

This can be used to implement logical branching, much like an if-else structure, although in most cases it does adversely impact the code's readability.

score 1 · Accepted Answer · answered Sep 29 '20 at 04:00

The code takes advantage of something called short-circuit evaluation, which isn't unique to JavaScript. In a broad sense, short-circuit evaluation means only evaluating a boolean expression if it is necessary to determine the ultimate outcome. It also takes advantage of the fact that in JS, assignment operators act as expressions.

Here's an example of what I mean:

let x;
console.log(x = 2) // 2

What this means is that you can have an assignment expression at the end of a list of conditions to act as an if statement. Due to short circuit evaluation, the assignment expression will only be evaluated if the ultimate true/false value of the full expression is not predetermined by the conditions.

let x;
(false && x = 1) // does nothing, false && ___ = false
(true || x = 2) // does nothing, true || ___ = true
(true && x = 3) // sets x = 3, true && ___ evaluates ___
(false || x = 4) // sets x = 4, false || ___ evaluates ___

score 0 · Answer 3 · answered Sep 29 '20 at 04:01

Nothing big changes in your code.

Heare simple syntax with the conditional operation in javascript var varname = condition ? true part code : false part code.

let fine = 0;
    const [actual, expected] = input.split('\n').map(item => {
        const [day, month, year] = item.split(' ').map(Number);

        return { day, month, year };
    });

    fine = (actual.year === expected.year && actual.month === expected.month && actual.day > expected.day) ? ((actual.day - expected.day) * 15) : 0;

    fine = (actual.year === expected.year && actual.month > expected.month) ? ((actual.month - expected.month) * 500) : 0;

    fine = (actual.year > expected.year) ? 10000 : 0;

    console.log(fine);

Something more

How does this code work as conditional statements, why aren't there any "if" or "else if" statements?

3 Answers3