How to use conditional(ternary) operator in javascript

Question

Given a string, I need to get the number of occurrence of each character in the string.

Input : "Hello world" 
Expected Output : { H: 1, e: 1, l: 3, o: 2, ' ': 1, w: 1, r: 1, d: 1 }

When I use if else condition the logic works fine , but not with ternary operator.

const string = "Hello world";
const chars = {};

for(let char of string) {
    if(!chars[char]) chars[char] = 1;
    else chars[char]++;
}
console.log(chars); // { H: 1, e: 1, l: 3, o: 2, ' ': 1, w: 1, r: 1, d: 1 }

But, When I replace the if else condition with ternary operator, the output is unexpected

chars[char] = !chars[char] ? 1: chars[char]++
console.log(chars); // { H: 1, e: 1, l: 1, o: 1, ' ': 1, w: 1, r: 1, d: 1 }

it does not make sense, to increment a variable and assign the value of the same variable to itself. — Nina Scholz
– Nina Scholz, Commented May 26, 2018 at 17:42

Jonathan Lonowski · Accepted Answer · 2018-05-26 17:54:26Z

3

In this case, you'll want to move the ++ before chars[char]:

chars[char] = !chars[char] ? 1 : ++chars[char]

Or just an addition:

chars[char] = !chars[char] ? 1 : chars[char] + 1

Or, even shorter:
chars[char] = (chars[char] || 0) + 1

Whether you place the ++ before or after a value changes the value it returns:

After (chars[char]++), the operator will return the original value, 1, as it increments to 2. The assignment operator, then, will receive the 1 and place it back into chars[char], undoing the increment.
Before (++chars[char]), the operator will return the modified value, 2, for the assignment to use. Here, the operators aren't in conflict with each other, as they're both setting the same value.

edited May 26, 2018 at 17:54

answered May 26, 2018 at 17:38

Jonathan Lonowski

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3 Comments

Scott Sauyet Over a year ago

Yes on the "even shorter" version, as it's also clearer!

Jonathan Lonowski Over a year ago

@Prem With your if..else snippet, you aren't using the value returned from the ++ operator. Without assigning it anywhere, that return value is discarded and has no effect on the results of the script. In switching to ternary, you also began to assign the result of ++, making it more relevant.

Prem Over a year ago

@ScottSauyet There was a typo in my comment in this post, but I couldn't edit the typo. Hence I deleted it. Your comment clarifies my doubt.

paul · Accepted Answer · 2018-05-26 17:35:42Z

1

You are setting chars[char] to the result of chars[char]++ which is 0

You want: chars[char] = !chars[char] ? 1: chars[char]+1

answered May 26, 2018 at 17:35

paul

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