Get the index of the first element in an array with value greater than x

Question

I have this array:

var array = [400, 4000, 400, 400, 4000];

How can I get the index of the first element with value greater than 400?

Note: While making sure this question was unique, I came across questions asking this same problem- but for different programming languages.
If there are duplicates of my question that apply to JS, I'd really want to see them.

Answer 1

You can use findIndex here

check this snippet

 var array = [400, 4000, 400, 400, 4000]; var index=array.findIndex(function(number) { return number > 400; }); console.log(index); 

Answer 2

You can use a simple for loop and check each element.

 var array = [400, 4000, 400, 400, 4000]; var result; for(var i=0, l=array.length; i<l; i++){ if(array[i] > 400){ result = i; break; } } if(typeof result !== 'undefined'){ console.log('number greater than 400 found at array index: ' + result); } else { console.log('no number greater than 400 found in the given arrry.'); } 

Read up: for - JavaScript | MDN

Answer 3

var array = [0, 0, 3, 5, 6];
var x = 5;
var i = 0;
while (array[i] <= x) {
    i++;
}