Console log not printing variable from function

Question

Trying to print the variable 'randomWord' to console.log, but chrome says it is not defined. It looks like it's defined to me. Why won't it print to the console.log?

function strt(){

//get random word from words[] array
var randomWord = words[Math.floor(Math.random()* words.length)];

var wordLength = randomWord.length;


//create a blank boxes or div elements for holding each letter of 
// selected random word
for(i = 0 ; i< wordLength; i++){

var divTag = document.createElement("div");
divTag.id = "div" + i;
divTag.className = 'wordy';
//divTag.innerHTML = randomWord[i];
hangManDiv.appendChild(divTag);

};// end for loop

//disable start button
document.getElementsByName("startB")[0].disabled = true;

return randomWord;

}//end strt()

console.log(randomWord);

Are you calling console.log outside the function ? the randomWord var is defined only in the scope of the function - you need to do this: console.log(strt()); — Adidi
– Adidi, Commented Mar 29, 2013 at 0:02
The variable is defined but it is local to the function strt(). Move the console.log to before the return statement. — andyb
– andyb, Commented Mar 29, 2013 at 0:03

JCOC611 · Accepted Answer · 2013-03-29 00:03:08Z

5

The variable randomWord is out of the scope. You define the variable inside a function, and then call it outside of it.

You should either define the variable out of the function or call it inside of it:

function strt(){
   var randomWord;
   ...
   console.log(randomWord);
   return randomWord;
}//end strt()

Or

var randomWord;
function strt(){
   ...
   return randomWord;
}//end strt()
strt(); // Call the function
console.log(randomWord);

For the latter, consider that randomWord won't have changed when JS executes the console log function; therefore, it will be null. In other words, you must call the function before you log it.

answered Mar 29, 2013 at 0:03

JCOC611

19.8k15 gold badges71 silver badges90 bronze badges

Sign up to request clarification or add additional context in comments.

6 Comments

ErikE Over a year ago

"The variable" what? Specificity would be an improvement. :) Also, your second code block needs to show strt();. A third, possibly better, option than declaring randomWord outside the function would be console.log(strt());.

JCOC611 Over a year ago

@ErikE: True, although I'll leave it as it is because it illustrates the point better, as to not confuse others. For example, if one wanted to console.log more than one variable...

oxxi Over a year ago

Ok, so I ended up putting the console.log inside the strt() function. It works now. One question though. In the strt() function I'm returning the 'randomword' variable. Why can't I use 'randomWord' outside the strt() function if it's being returned.

JCOC611 Over a year ago

@user1965121: Think of it as a sequence of steps. In your case, the console log occurs before the variable is returned. Additionally, for it to work, the variable you make reference to must be in the scope of the call. In other words, you would need to have something like newvar = strt() and you would make reference to this new variable rather than the initial one.

ErikE Over a year ago

@user1965121 a return value from a function completely disappears if you don't store it into a variable. It doesn't automatically create a window-level/global-level variable of the name randomWord. Consider that in your function you could just as easily do return 'the word is ' + randomWord + '.';--how could this create a variable in the outer scope and what name would it have? Instead, you need to do var wordresult; wordresult = strt(); console.log(wordresult); to get what you're expecting: a variable with the result of the strt function's execution as its value.

|

Collectives™ on Stack Overflow

Console log not printing variable from function

1 Answer 1

6 Comments

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

6 Comments

Linked

Related