Update js codes.docx

Author: priya42bagde

js codes.docx

@@ -3040,7 +3040,46 @@ Output: 1
 However, if you use the assignment operator for a reference value, it will not copy the value. Instead, both variables will reference the same object in the memory:
 
  JS objects (as non-primitive data types) differ because they have reference values and those values are mutable. This means they share a memory address when shallow copied.
- 
+=========================================================================================================================================================================
+Hoisting:
+var x = 1;
+
+function greeting(){
+  console.log("Hi greeting");
+}
+
+var greeting1 = function (){
+  console.log("Hey greeting1");
+}
+
+var greeting2 = () =>{
+  console.log("Hello greeting2");
+}
+
+console.log(x); //1
+greeting(); //Hi greeting
+greeting1(); //Hey greeting1
+greeting2(); //Hello greeting2
+-----------------------------------------------------------------------
+console.log(x); //undefined
+greeting(); //function body will assign before executing //Hi greeting- as output post execution
+//greeting1(); //Uncaught TypeError: greeting1 is not a function because its treeting like a variable not a function, so rather then storing a function body it's storing undefined by default.
+//greeting2(); ////Uncaught TypeError: greeting1 is not a function because its treeting like a variable not a function, so rather then storing a function body it's storing undefined by default.
+
+var x = 1;
+
+function greeting(){
+  console.log("Hi greeting");
+}
+
+var greeting1 = function (){
+  console.log("Hey greeting1");
+}
+
+var greeting2 = () =>{
+  console.log("Hello greeting2");
+}
+==========================================================================================================================================================================