JavaScript function is not working as per requirement

Question

As per learn and try I am working on very simple function to understand features of JavaScript function. But after create function I am getting some issues, I hope someone can help me for my following questions.

Questions are:-

a) Why console.log from browser is showing ReferenceError: c is not defined ?
b) Why p id="demo" is not able to show result ?
c) Why alert(c); outside of the function is not showing result once browser load/refresh?
d ) Why return(c); is not working?

function show(a, b){
  
  var c = a+b;
  alert(c);
  console.log(c);
  return(c);
  
}
alert(c);
document.getElementById("demo").innerHTML = c;

<p onclick="show(10, 20)" >This example calls a function which performs a calculation, and returns the result:</p>

<p id="demo"></p>

You should not get that reference error from the `console.log` line, but from the second alert statement. — Bergi, Feb 19 '15 at 11:17
http://stackoverflow.com/q/14028959/1048572 gives the answer to your last question. Also, there is no `c`; where should it come from, your function has not even been executed yet? — Bergi, Feb 19 '15 at 11:19

score 2 · Answer 1 · answered Feb 19 '15 at 11:17

2

Because function braces in javascript define scope and all variables declared inside it aren't visible outside of it. So c is undefined where you're alerting it.

function show(a, b){
   var c;
   //'c' is visible only in the function
}

answered Feb 19 '15 at 11:17

kidwon

4,080
4
25
44

score 2 · Answer 2 · answered Feb 19 '15 at 11:19

As i say in my comment :

If you define c in function is only define in function

You can do this :

HTML

<p onclick="show(10, 20)" >This example calls a function which performs a calculation, and returns the result:</p>

<p id="demo"></p>

JS

    function show(a, b){

      var c = a+b;
      document.getElementById("demo").innerHTML = c; 

    }

score 1 · Answer 3 · answered Feb 19 '15 at 11:17

1

Variable c is local to function show. So when you call it outside of the funciton,

you get the error ReferenceError: c is not defined and as c is nothing, p demo doesn't show anything

answered Feb 19 '15 at 11:17

snehatulsi

231
1
11

score 1 · Answer 4 · answered Feb 19 '15 at 11:20

1

    function show(a, b){
      
      var c = a+b;
      return(c);
      
    }
    var d = show(10,20);
    document.getElementById("demo").innerHTML = d;

    <p onclick="show(10, 20)" >This example calls a function which performs a calculation, and returns the result:</p>

    <p id="demo"></p>

answered Feb 19 '15 at 11:20

snehatulsi

231
1
11

thanks for your suggestions. can you please help me to learn more about return()? – Sandy Feb 19 '15 at 12:38
return is a statetement, when executed returns from the current function. The lines after that are never executed. And we can return values in the return statement. We can return both data/function in javascript. – snehatulsi Feb 19 '15 at 16:44

score 0 · Answer 5 · answered Feb 19 '15 at 11:31

Try This:-

        var c = "10";
        function show(a, b){
  
            c = a+b;
            alert(c);
            console.log(c);
            document.getElementById("demo").innerHTML = c;  
        }       
        window.onload = function () {
            document.getElementById("demo").innerHTML = c;
        };

<p onclick="show(10, 20)">This example calls a function which performs a calculation, and returns the result:</p>

    <p id="demo"></p>

JavaScript function is not working as per requirement

5 Answers5