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JavaScript JavaScript Loops, Arrays and Objects Tracking Multiple Items with Arrays Iterating through an Array

Posted by Robert Allred
Robert Allred
920 Points

What am I missing?

What am I missing here? Because to me this looks like it should repeat this process of printing to the log the first number that is in the array, which also removes it and then repeats that process.

script.js 
var temperatures = [100,90,99,80,70,65,30,10];
for ( var i = 0; i < temperatures.length; i += 1){
 console.log(temperatures.shift());
}
index.html 
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>JavaScript Loops</title>
</head>
<body>
<script src="script.js"></script>
</body>
</html>
Nick Gentile
Nick Gentile
4,237 Points
Nick Gentile
Nick Gentile
4,237 Points

I'm not sure what you are doing with .shift() -- but I can tell you that is where you are wrong.

What you want to do is log the value of temperature, at the given index point in the for loop (i).

Remember that to log a value at a certain index point of an array we do something like this: 

array[index]

Big hint if you are still struggling ::

Your array is called TEMPERATURES

and the variable you are using to grab the index is [i]

1 Answer

Charles Wanjohi
Charles Wanjohi
9,235 Points
Charles Wanjohi
Charles Wanjohi
9,235 Points

shift() method removes the first element in an array.Therefore every time the shift() method is executed, the length of the array is reduced: After executing the loop for 4 times, the length of the loop will be 4 because the first 4 elements will have been removed already.Therefore the (i < temperatures.length) condition in the for loop will not be satisfied i because the value for i is 4 by now .As a result the loop terminates . You can achieve the desired result by implementing the code as follows:

var temperatures = [100,90,99,80,70,65,30,10];
while ( temperatures.length>0){
 console.log(temperatures.shift());
}

Hope this helps you