Inconsistent results of using indexOf on an array in an if-statment

Question

I am in the process of learning node. The following code is giving me inconsistent result ie, if i give argv1.indexOf('test'), then the code is unable to find the text, but, some time it is returning true. Why is it so.

function process(argv1) {
    if(argv1.indexOf('test')) {
        console.log('Text is available in array.');
    }
}
process(['test','one','help', 'one', 'two']);

What would you expect indexOf to return in this case?

Felix Kling
– Felix Kling

2014-12-09 01:18:56 +00:00
Commented Dec 9, 2014 at 1:18 — Felix Kling
– Felix Kling, Commented Dec 9, 2014 at 1:18
Did you bother to read the documentation for indexOf?

user663031
– user663031

2014-12-09 03:12:15 +00:00
Commented Dec 9, 2014 at 3:12 — user663031
– user663031, Commented Dec 9, 2014 at 3:12

volter9 · Accepted Answer · 2014-12-09 01:49:31Z

5

That's because indexOf returns an index of matched element. If element not found, it will return -1.

What you need is to change the condition:

function process(argv1) {
    if(argv1.indexOf('test') !== -1) {
        console.log('Text is available in array.');
    }
}
process(['test','one','help', 'one', 'two']);

Edit: As pointed out @Havvy, in case of test, .indexOf will return 0 which would be casted to false. For other array elements, their indices would be converted to true, as any non zero numbers would be casted to true. More about javascript evaluation you can read here.

edited Dec 9, 2014 at 1:49

answered Dec 9, 2014 at 1:14

volter9

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2 Comments

Havvy Over a year ago

Furthermore, if will typecast the result to a boolean. The value 0 gets typecasted to false, so ['test', 'foo', bar'].indexOf('test') evaluates to 0, and if (0) {} evaluates to false, not executing your block.

volter9 Over a year ago

Yeah, of course, and if the element index would be more than zero, it would cast to true.

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