The reason it returns undefined
is because the a
inside function foo() has a local scope right?
It does not have any co-relation with the a
in the global scope which has a value of 1 , right?
function foo(){a=a+1;}
var a = 1;
foo();
The reason it returns undefined
is because the a
inside function foo() has a local scope right?
It does not have any co-relation with the a
in the global scope which has a value of 1 , right?
function foo(){a=a+1;}
var a = 1;
foo();
No it returns undefined because you are not logging anything
how do I make it returns two then?
this does not work :
function foo(){console.log(a+1);}
var a = 1;
foo();
That logs 2
. If you want to return 2, you can just use return a+1
.
I see thanks! (20 character)
Mark this as solved now?