JS Question

Question

var x = 2;
var y = 4;

if((y > x || y++ >= 4) && ++y === 5){
x = 1;
} else{
x = 4;
}
console.log(x); // 1
console.log(y); // 5

In this Task, why final value of y isnt 6? Doesnt y increment by 1 after first comparison, since it is true and then increments again by 1 in second comparison, resulting in y = 6 and therefore x = 4

Answer 1

When either operand of logical OR (||) is truthy-value it evaluates to that truthy-value.
In short it returns first truthy operand.

Thruthy values:
https://developer.mozilla.org/en-US/docs/Glossary/Truthy

If first operand of || is truthy second operand is not checked.
This is known as "Short Circuit Evaluation". In your case JS doesn't evaluate second operand y++ >= 4 as y>x is true.

https://codeburst.io/javascript-what-is-short-circuit-evaluation-ff22b2f5608c

example:
console.log({ } || [ ]); // object literal is truthy. || returns { }

console.log("" || "o"); // empty string is falsy. returns next value "o" which is truthy.

console.log(false || null || true || {}); //returns true, the first truthy operand in expression.

This can be useful when you want to use some default value.

let name = prompt() || "Anonymous";
alert(name);

if you press cancel prompt returns null (falsy) in that case - || returns "Anonymous".

Answer 2

Thanks Omkar for your great explanation and for the sources, It really helped me see the bigger picture. Also thanks lpang for the sources.

Answer 3

What we see here is an effect of short circuit evaluation. You will find answer plus explanations in the following

https://en.m.wikipedia.org/wiki/Short-circuit_evaluation

https://www.tutorialspoint.com/short-circuit-evaluation-in-javascript