foreach,map,filter,spread,rest,destructuring

Mehak-Kamran · Mehak-Kamran · commit 87912f26322f · 2025-01-03T19:38:11.000Z
diff --git a/revision/rev4.js b/revision/rev4.js
@@ -0,0 +1,98 @@
+//for each
+
+let names=['alice','bob','peter','jane'];
+
+names.forEach((name)=>{
+    console.log(name+" ji");
+
+})
+
+//map
+let newnames=names.map((name)=>{
+   return name;
+})
+
+console.log(newnames);
+
+//filter
+let condnames=names.filter((name)=>{
+   return name==='alice'
+})
+console.log(condnames)
+
+//destructuring
+let obj={
+    name:"marry",
+    age:[20,30,40],
+    address:{
+        karachi:"gulshan",
+        lahore:"laal qila"
+    },
+    country:"Pakistan"
+}
+
+let access1=obj.age.at(0)
+let access2=obj.address.karachi
+console.log(access1);
+console.log(access2);
+
+let {karachi,lahore} =obj.address
+console.log(lahore);
+
+let [f,s] =obj.age
+console.log(s)
+
+//spread
+//premetives pass by value
+let a=4;
+let b=a;
+console.log(b+1);
+console.log(a);
+
+//nonpremitives pass by ref
+let array1=[1,2,3]
+// let array2=array1
+
+// array2.pop()
+// console.log(array2);
+// console.log(array1);
+
+//to solve this prob we use spread op
+let array2=[...array1];
+array2.pop()
+console.log(array2);
+console.log(array1);
+
+//same for object
+let obj1={
+    name:"hello",
+    color:"black"
+}
+let obj2={...obj1}
+obj2.name="mak";
+console.log(obj1)
+console.log(obj2)
+
+//spread op is only level one operator (que asked in interview)
+
+let list={
+    add:{
+        name:"mak"
+    },
+    tag:"sad"
+}
+let list2={...list}
+list2.add.name="fak";
+console.log(list)
+console.log(list2)
+
+//rest op
+//bajay ye k ap multiple var banay in fun par we can store val in an array code optimization and memorysaves wrt space 
+
+let bilal =function(...val){
+console.log(val)
+}
+bilal(1,2,3,4);
+
+
+
