Why does it say my variable c is undefined when I returned it?

Question

https://code.sololearn.com/Wa63fJ8TTbGC/?ref=app

Answer 1

The main reason is because variable c is a local variable of the function and cannot be invoked outside the function. This is how I altered your code:

function hippy(one, bleh)
{
var c = (one+ "is better than "+bleh);
return c;
}

str = hippy("js", "c++");
alert(str);

console.log(str);

Now, we return the local variable c to a global variable str, and console.log the global variable.

Answer 2

@AceDev:

You seems to have modified your code by following @Hatsy Rei advice... but you do too much: it's not necessary to do same with your two other functions ;P

function hi(){
/*
var a= alert("howdy") ;
return a;
*/
alert("howdy") ;
// is enough, as your function call expect only a behaviour, not a returned value
}
hi(); // by calling this function, you will just alert("howdy") you doesn't store a value if one is returned ;)

function foo (boo){
// same here
/*
var b = alert(boo + "is fun!") ;
return b;
*/
alert(boo + "is fun!") ;
}

foo("Js ");

function hippy(one, bleh){
var c = alert(one+ "is better than "+bleh) ;
return c;
}
hippy("js", "c++"); // with this you don't store the returned value
// and you still cannot access the 'c' variable... look better at @Hatsy Rei code:
// console.log(c);
var result = hippy("js", "c++"); // with this you store the returned value
// and you log it:
console.log(result);

But anyway, the returned result isn't really interresting: it's the returned value of alert() function, wich return nothing ^^
So, maybe you are attempting to rather do:

function hippy(one, bleh){
var c = one+ "is better than "+bleh ;
return c;
}
var r = hippy("js","c++");
alert(r);
// or shortener:
alert(hippy("js","c++"));