Thruthy IF ELSE statement

Question

Let a = 10;
Let b = 11;
Let c;

console.log(a + “” + b);

If (!(a++ >= 10 || b++ >= 11)) {
c = a + b;
console.log(a + “” + c);
} else {
console.log(a + “” + b);
}

// 1011
// 1111

Please can someone explain why I’m getting different output

Answer 1

Reason is because a in incremented inside if statement with a++, this means a = a +1 after this code but not inside this line.
So if(!(a++>=10 || not important because first is true)) - here a is 10
After this a is 11
Thats why only a is changed not b, and you got 1111 as output
a=11,b=11

Answer 2

a++ increment value of variable after this line of code.
Or (||)will return true or false, but if first is true it wont even check second part, because value will be true.

Answer 3

Thanks Sanja Panic. That was really helpful

Answer 4

Ipang thank you.

Answer 5

Just additional resources

https://jay-cruz.medium.com/short-circuit-evaluation-667d01b54172
https://medium.com/khojchakra/understanding-short-circuit-evaluation-in-javascript-31b5770a41f7

Answer 6

No problem bro 👌

Answer 7

let a=10;
let b=11;
let c;

if(!(a++>=10) || (b++>=11)){

c=a+b;

console.log(`${a}${c}`);
}

else{

console.log(`${a}${b}`)
}