Variable undefined in Java script

Question

function getForm()  
{   
  var arry =[ [0,1] ,[1,2],[2,3],[3,4]];         
  var a;     
  var x = 1 ;        
  console.log(x);     
  for (i = 0; i < arry.length; ++i)      
  {         
    if (arry[i][0] == x)          
    {             
      a = arry[i][1];          
    }         
    console.log(a);  
  } 
}

I am totally new to JavaScript, I was running this programming in Firebug. I have got the output as undefined I was expecting the output x=1,a=2(I mean first console.log and second one). I was defined all the variables inside the function only.

You just define a function and never called it. So, firebug after declaring functions, says undefined — Super Hornet
– Super Hornet, Commented Mar 4, 2015 at 5:35
I have got error in the 3rd line , It was taking array in the for loop as an normal text not in blue color — hackwithharsha
– hackwithharsha, Commented Mar 4, 2015 at 5:44

user229044 · Accepted Answer · 2015-04-26 00:13:31Z

2

Your mistake is console print outside the if condition, so first time condition going false (return undefined) and second time condition satisfied (then print correctly),

you should move console into if condition,

<script>
    var arry =[ [0,1] ,[1,2],[2,3],[3,4]];         
    var a;     
    var x = 1 ;        
    console.log(x);     
    for (i = 0; i < arry.length; ++i){         
         if (arry[i][0] == x){        
            a = arry[i][1];          
            console.log(a);  
         }         
    } 
</script>

Console Result:

 1
 2

View on jsFiddle

edited Apr 26, 2015 at 0:13

user229044♦

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answered Mar 4, 2015 at 5:39

Jaykumar Patel

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3 Comments

hackwithharsha Over a year ago

Yeah, It was working fine !! If I declare array outside the function , I am not able to access it !!

M3ghana Over a year ago

But why did not the x print initially? I mean this var x = 1 ;

Jaykumar Patel Over a year ago

@Clouds off-cause we print x initially but OP have some purpose to not print initially.

Felix Kling · Accepted Answer · 2015-03-04 05:35:33Z

0

You are logging a in each iteration, but a is not set until the if statement is true, which only happens in the second iteration.

Lets go through it step by step:

// 1. iteration
i = 0
arry[i][0] == x => false // 0 == 1
console.log(a) // undefined

// 2. iteration
i = 1
arry[i][0] == x => true // 1 == 1
a = arry[i][1] // a = 2
console.log(a) // 2

// 3. iteration and beyond
// the condition is always false, so `a` keeps its value
i = 2
arry[i][0] == x => false // 2 == 1
console.log(a) // 2

answered Mar 4, 2015 at 5:35

Felix Kling

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Comments

David · Accepted Answer · 2015-03-04 05:36:08Z

0

It outputs undefined because the first time you call console.log(a) the variable a is undefined.

You declare a without a value:

var a;

Then in the first iteration of the loop you check for this:

if (arry[i][0] == x)

When i is 0, arry[i][0] is also 0. And you defined x as 1. 0 does not equal 1, so this condition is false. Which means the code inside the conditional block, which sets a value to a, doesn't execute. But then the code after it does.

At that time a is undefined, so console.log(a) logs undefined.

answered Mar 4, 2015 at 5:36

David

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Comments

super_user · Accepted Answer · 2015-03-04 05:40:11Z

0

console.log(x) is working fine

in first interation , the if condition is false , so 'a' will not get initialized and hence in console you will get undefined , next iteration onwards you will get the desired console output,

if you are running on firebug extra 'undefined' will be displaed because calling a function which does not return anything will result 'undefined' on the console

answered Mar 4, 2015 at 5:40

super_user

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Comments

gettingLearned · Accepted Answer · 2015-03-04 05:56:20Z

0

You declared the function but never invoked. Invoking the function will give you the desired results, since when I tried to execute the statements without function wrap, i.e

    var arry =[ [0,1] ,[1,2],[2,3],[3,4]];
    var a;     
    var x = 1 ;        
    console.log('Value of x', x);     
    for (i = 0; i < arry.length; ++i) {         
       if (arry[i][0] == x) {             
          a = arry[i][1];          
       }         
       console.log('Value of a in ', i , 'iteration' ,a);  
    } ;

this was the output:

Value of x 1

Value of a in 0 iteration 2864415817892.274

Value of a in 1 iteration 2

Value of a in 2 iteration 2

Value of a in 3 iteration 2

edited Mar 4, 2015 at 5:56

answered Mar 4, 2015 at 5:46

gettingLearned

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