Ternary operator not working properly in JS

Question

I am running the following code and getting unexpected results:

var a = 1, b =2 ,c = 3;
console.log(5*a+ b>0?b:c);

Expected result was : 7 but getting 2.

It's because it's returning b. You should use brackets to separate the ternary — Kobe, Nov 01 '19 at 11:32
@AvinashKumar why would you get *a string*? None of the operands is a string, so you won't get concatenation. — VLAZ, Nov 01 '19 at 11:33
Your condition is `5 * 1 + 2 > 0`. If this is true (which it is), then `b` is returned (`2`). Works as coded. — deceze, Nov 01 '19 at 11:34
Note that JavaScript doesn't care that you put the space where you did — the space doesn't do anything for operator precedence, only operators and parentheses do. — Amadan, Nov 01 '19 at 11:35
It's for strings but has the same setup (`+` vs. `...?...:...`): https://stackoverflow.com/questions/12847853/using-the-ternary-operator-with-string-concatenation — Andreas, Nov 01 '19 at 11:37
console.log(5*a + (b>0?b:c)); Just check priority on mdn docs — Shubham Verma, Nov 01 '19 at 11:57
You can use [AST Explorer](https://astexplorer.net/#/gist/05149f428a64055b822c2db39d0dfce1/bae294ea9b3e3383e40ba2749f3dc816041aadf3) to check the order of execution of each operator — adiga, Nov 01 '19 at 11:57

Kobe · Accepted Answer · 2019-11-01T11:45:51.477

Your code has the right concept, but wrong execution. The ternary is doing its job properly.

At the moment, your code is executing like this:

const a = 1
const b = 2
const c = 3

// This will evaluate to true, since 5 * 1 + 2 = 7, and 7 is greater than 0
if (5 * a + b > 0) { 
  // So return b
  console.log(b)
} else {
  console.log(c)
}

You should use brackets to separate the ternary:

const a = 1
const b = 2
const c = 3

console.log(5 * a + (b > 0 ? b : c));

Ternary operator not working properly in JS

1 Answers1