Ternary operator not working properly in JS [duplicate]

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
– Kobe, Commented Nov 1, 2019 at 11:32
Your condition is 5 * 1 + 2 > 0. If this is true (which it is), then b is returned (2). Works as coded. — deceze
– deceze ♦, Commented Nov 1, 2019 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
– Amadan, Commented Nov 1, 2019 at 11:35

Kobe · Accepted Answer · 2019-11-01 11:45:51Z

5

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));

edited Nov 1, 2019 at 11:45

answered Nov 1, 2019 at 11:33

Kobe

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