fnOriginal = function(){
console.log("hello");
return 10;
};
var fun = new Function("fnOriginal()");
console.log(fun()); // prints undefined.
console.log(fnOriginal()); // prints 10
How would I make fun() return and print 10 like fnOriginal()?
Obviously,
var fun = new Function("return fnOriginal()");
new Function("code") is the same as function() { code }. If the code's missing a return statement, the function won't return anything.
Use following
fnOriginal = function(){
console.log("hello");
return 10;
};
var fun = new Function("return fnOriginal();");
alert(fun());
It will return 10.
