Lecture 7- Classes and Objects in Java and Solved problems

[Music] [Music] so welcome to lecture 7 classes and objects so in the last class as we had seen a program right multiplication of two complex numbers all right using class and objects so there we had a keyword called Static all right so what is the difference between static variable and the ordinary variable all right so when I'm talking about ordinary variable it is limited to certain scope right so which is well defined whereas when I'm talking about the scope of a static variable all right so that extends throughout the program we'll see with an example all right so that you can understand in a better way so here you go right so for example here I have let us say a function all right so it's a simple program right so here you have a c program all right so int main I I take it as a c you can change it to C+ plus also not a problem so int main right so here you have a function right so int function right whose return type is integer so int count is equal to zero which is not very difficult count equal to zero and then line number 12 I am using count Plus+ all right and then I have return count I take an example of a procedural oriented language C all right so you can just put C out when you put C out that's all that is called C++ now I'm calling the function all right when I'm calling the function you can tell me what will happen all right so function when I call so count is zero count Plus+ so I can expect the output what one good so you'll get the output one all right and next again I'm calling all right the new line again I'm calling the function so what will happen it goes count zero count Plus+ count is zero count Plus+ return count all right 0 0 + 1 is 1 so I have to get two ones as output all right so I have to get two ones as output so you can see that all right so I'll get two ones as output okay so what I will do shall I change this variable all right with stating we'll see what will happen all right so when I change this to static so let us see what is happening yes same program I have put static so that you can understand all right those who are learning first time about static variable this program will give you an idea so now as usual I'm calling the function so first one count zero count Plus plus so now count is one all right so here I'm writing count is one all right count plus plus count is so return count so it will return one all right so now when I use this variable right static variable the keyword static now my count is one that is why I separately wrote last time I did not write anything all right so now my count is one so again when I am calling the function in line number five all right in line number five when I am again calling this function static int right your count is one count Plus+ so now you will get the output two all right so this is the output I can expect return count so this is the output I can expect all right so you can see that when I run this code I can get the output like this so this is the difference between a simple variable on static variable so this program these two programs will be be useful to understand what is the meaning of the key word static so if you look at the last class we have use the keyword called Static now you understand based on these two programs you can understand so without static variable right line number 11 if I don't use static I am getting output as one one if I use static all right so count will be considered as one before going to line number five right and then when I it is calling the function again the count Plus+ 1 + 1 is becoming two and here we are getting the output one and two right so this is how the static variable is being defined so now the same complex number problem right the multiplication of two complex number problem so let us consider using we will write a program on Java all right so here you go so these are all the utilities I am importing in Java now I create almost the same the syntax is almost same few things here and there there will be a change so class complex you have double real and double imaginary all right you have two member data or data member and then set values the same like what we had seen in C++ set values and here the multiplication right so here you can't have the reference operator in Java you can't have the reference operator and star operator all right when I Define complex C1 all right so C1 is an object so I can print the identity hashmap all right so suppose I print let us say system.out.print and C1 I will get the identity Ash map right whereas in the case of C++ you have to slightly use the address operator slightly change the syntax I mean to say so you will have the address operator so here complex C1 and here you have complex C2 and you're having the result which is an object whose class is complex right so dynamically allocating a memory using new operator and this we are going to study right this syntax you please note this is nothing but the Constructor all right so line number 22 you are dynamically ating memory for result dynamic memory allocation complex result you put new operator so dynamically allocate the memory and you have complex followed by the function notation you call it as Constructor all right so now as usual multiplication of two complex numbers what we had seen in C++ the same syntax you are writing and return result all right so few things are changing so here we are not using the address operator like number 21 whereas we have used if you recall we use the address operator all right and complex result so here you go this is right the new syntax in Java right so few things here and there there will be a change and then display right the vo display same member function you call it as method right the same member function what we have used in C++ so here it will be the method and the method will give you the real plus I into imaginary all right and then I go to the main program all right so main program so it's a driver class the driver class is main because you have the main almost like in C++ right int main so here the syntax is static you know the meaning of Statics void main so this we'll see slightly later right string arguments string of arguments all right so now this function this inbuilt one is for taking the input from the user all right so this is the syntax that you are writing that means you are going to take some input from the user in fact line number 47 and 48 we are taking the input from the user for this so this inbuilt statement is required in line number 39 let us create two objects right let us instantiate two objects C1 and C2 C1 whose class is complex and then here you use a new operator right and then you use the Constructor and then another object so dynamically you are allocating memory for C1 and C2 that is a meaning okay so now enter your real and imaginary right enter your real and imaginary or the first complex number so that you are taking the input from the user this is what you use the scanner all right so it's like a scanner for scene all right SC do next Double next int that means the integer data type next Double so so this function the inbu function using your object SC under the class scanner right so here you go you are inbuilt all right this SC and scanner they are all the inbuilt class all right so which is useful this SC object is invoking next Double function next Double method inil method so that means whatever the user is entering will be stored in real one similarly next line line number 48 right this SC the object is invoking the inil method next Double it will be assigned to imaginary one all right so you are taking the input from the user user and then here C1 is invoking set values exactly like what we had studied in C++ okay so real one you are passing imaginary one so assume that it is 2.0 + 3.0 I something like that so real 1 is 2.0 and imaginary one is 3.0 similarly the second all right let us say 4 + 5 I 4.0 + 5.0 I so real two 4.0 and IM inary 2 right 5.0 so take the input from the user and set values so then the result method right the result object who class is complex all right so here you have so here the complex is invoking right the class complex so which is invoking the multiply Method All right so multiply method is being invoked here you are passing C1 and C2 as the arguments exactly like what we have studied in C++ almost similar all right so the complex so it is more or less that it is like as if you have created some objects so that object is invoking multiply so this syntax is also valid or you create some complex let us say C3 all right so that C3 will invoke the multiply that is also possible all right or otherwise simply you put the class name of the class do multiply C1 comma C2 right so then the resultant will be stored in result that is the meaning then result of the multiplication so result. display will show result do real and result do imaginary right and then SC do close will close the file because you are taking the input using SC and then you do SC do close right this object is invoking the close it's a kind of file closing right so now when I run the code I use assume that even if you put two three it is assumed that you put 2.0 and 3.0 4.0 and 5.0 same input we will give and you can see the result of multiplication you will get this is the output okay so how to create a class right and how can you define a class and how to create objects right all these we have studied with the help of C++ as well as a Java program also now you are familiar with the static keyword right so we will see one more case study right one more problem so write a C++ program to add two distances where feet and inches are given right so you are given a feet let us say 9 and seven all right another one is8 and six all right so these is feet and these are all inches so if you given this feet and inches I want the output of addition right so 12 in is 1 ft so 7 + 6 13 that means 1 ft will be added 1 + 8 is 9 9 + 9 is 18 so this I have to get the output on 1 in right when I'm adding these two I have to get the output 18 ft and 1 in all right so this how do you do with the help of object oriented programming you can think of so how do you think you can create a class distance and then you have to include feet and inches as the object and then you have to think of right geter and Setter values display values right and then the main program you have to write addition of two distances all right so this we will see so let us consider I have a class distance all right so include IO stream because I'm going to use C in C out so therefore using name space STD so class distance the name of the class is distance I am using fet and Ines right I'm using feet and inches so here I I have set values set values in F and in I in F and in I so feet is equal to F so whatever you are taking the input that is being passed it will be stored as feet and inches under that particular object so feet is equal to F and inches is equal to I all right so now here you can see I have ADD name of the function all right name of the member function and I have an object D whose class is distance and the return type is also distance and here I have distance result so that means I will have to right I need a two distances all right so for example D1 + D2 which I'll store in D3 resultant distance so now I can use iier all right without two arguments in the last program in the multiplication of two complex numbers we use two arguments so can you write it in one argument yes you can do that what it will do one object will invoke this add function all right you will have like this D1 do add and then you're passing D2 all right you will call similar to this so that means means here you have inches simply inches you're not having D1 do inches simply inches and simpol simply feet so this will be taken care that is nothing but D1 do in D1 do feet all right so one object's inches will be added with another object's inches all right so for example I have 97 86 right so first what I do I'll add the inches so here I will add 7 + 6 what will happen 13 all right so here I have to use some logic when it is going Beyond 12 all right I have to increment my feet all right so that we will do that so inches you are adding first so we are getting result. inches is right now 13 assume that right then what I will do I'll add the feet 9 + 8 feet plus d do feet the object that you're are going to invoke so that is going to be your 9 and rest is your passing 8 all right so 9 + 8 correct so you will have how much 17 and then so here you have result do Ines integer division 12 so 13 by 12 integer division what it is 1 all right so one will be added feet plus D1 do feet and result. in /2 divided by 12 13 by 12 integer Division 1 so 17 + 1 so right now it will be 18 and then what you have to do you have to do mod all right so 13 mod 12 13 mod 12 is what 1 your result dot inches will be 1 right your result dot Ines will be 1 13 M 12 so 17 + 1 18 13 mod 12 is 1 so this will be my resultant for this test cases this test case Okay so this you are returning as result this you are returning as result so why display you put feet and inches why display you put feet and inches so let us see because it is slightly different code what we have studied in the case of addition or multiplication of two complex numbers all right so here we'll see how we are going to use this syntax in the main program so you create three objects one is two distances D1 plus d D1 plus D2 put it in result right so basically I want like this result equal to D1 + D2 all our objects so D1 do in D1 do feet D1 do feet D1 do in D2 feet and D2 do in so resultant you will get result do feet and result do Ines okay and then you have set of variables Right feet inches feet 1 inches one feet 2 and inches two okay so now you are entering feet one and inches one so assume that I entered 9 and 8 all right I entered 9 and 8 so you are setting the values feet 1 Ines 1 that means D1 is invoking set value right so the set value member function here you go feet and inches will be set okay feet is equal to F and inches is equal to I so 9 and 8 whatever right so 9 and 8 and second one I'm giving 8 and six all right so your Fe entering your fe2 is eight and and inches 2 is six fe2 and inches 2 so you are setting the values so this will be set so this is for D1 and this is for D2 because D2 is invoking here all right set values will be called so D2 do feet is 8 and D2 do Ines is 6 so now this is what the syntax I wrote If you recall right so d D1 do add D2 exactly I wrote this right so D1 do add D2 you are putting whatever be the result you are putting as the result that is the object right so result object is under class distance okay so now what is happening D1 is invoking the member function add by pausing D2 all right so that means you pausing what 8 and six you're pausing 8 and six so when I'm pausing 8 and six all right so here your D do inches is what 8 right inches feet six right d do inches is six 8 and six so here I change 98 and 86 all right what about my simple inches I don't have any object dot inches all right so which is nothing but this inches 8 so that means here what is happening 8 + 6 will be added all right so here it is nothing but 8 + 6 will be added so when 8 and six is being added it is 14 right this is what exactly happening 8 and six is nothing but 14 right D1 is invoking you may ask the question how this D1 do Ines is going right so D1 is invoking since D1 is invoking so here the simple inches plus d do in D do Ines we know it is 8 inches is six right so 8 + 6 is being added and it is nothing but 14 now you add feet right so feet is nothing but 9 and 8 so 9 and 8 will be added along with result. in /2 so here I have 14 14/2 is 1 right so 9 + 8 17 17 + 1 18 18 ft I got still inches I have not got all right so inches is nothing but 14 mod 12 14 mod 12 is what 2 all right so that means 18 ft 18 ft 2 in I get as the output right when I'm adding this 18 and 2 14 mod 12 is right this is what we are telling 14 mod 12 is line number 22 you look at line number 22 14 mod 12 is 2 so we'll get two as the output so totally you will have 18 ft and 2 in as the output all right so this we are displaying result. display we are doing all right so let us see what are the test cases test cases assume that yeah I give another test case 159 15t 9 in 18 ft and 7 in when you add this all right so 9 + 7 I will get 16 15 + 18 I will get 33 so 33 + 1 will be added 16 mod 12 will which will be 4 16/2 is 1 so 33 + 1 I will get 34 16 mod 12 I'll get 4 so 34 ft and 4 in I have to get as a output so if you look at all right so I'm getting exactly the whatever the expected output we are getting this as the output 34 ft and 4 in I hope it is clear this problem is clear can we write the same program in Java right so almost same so let us try so earlier what you can do you try to write it in both C++ and Java and then later at one point of time we'll stick only C++ okay so here you have a class distance feet and inches and here you have a member function same member function what we have studied in C++ what we have seen in C++ set values all right so similar right so whatever we have done same thing you are doing distance is a return type all right so return type is an object distance add distance D all right so D is an object you have a distance so returning object is possible these two programs whatever program that we see saying in C++ as well as in Java the return can you return the object right so these two programs are telling yes you can return the objects and here you go you have a result object whose class is distance dynamically you are creating a memory space with the new operator and distance is nothing but the Constructor so rest is similar you are adding inches plus d in D do in whatever we have studied in C++ similarly feet plus d do feet and then result. in /12 integer division so whatever be the resultant you are putting over here result do feet similarly result do Ines is equal to result do in mod 12 the reminder return result right so almost the similar syntax so one here line number 20 there is a change all right so dynamically you are creating a object and vo display so display you are displaying feet and inches so this is also similar and then if you look at the class you are not putting any semicolon the class ends in line number 32 and here you go I may use certain input therefore I have an object SC under the inbu scanner inbu class scanner all right your program starts here the class main you call it as driver class all right you have a main program starts from here so you are creating three objects all right so D1 and D2 all right dynamically creating a object and the third one is for result right so D1 D2 new operator Constructor D2 new operator Constructor and then you have result all right so feat one all right so feat one is taking the input whatever input you are giving or input that you are getting from the user that will be stored in feet 1 line number 45 and similarly whatever the input you are getting from the user you are storing in inches one right so feet one and inches one that is your first object all right first objects feet one and inches or feet and inches simply so feet one and inches will be set all right so the object D1 is invoking set values member function all right so feed one and Ines one whatever you are pausing all right so that will be set for D1 feet and D1 do in right so this is what exactly will happen if you look at line number 14 and 15 D1 do F feet and D1 do Ines will be set similarly for the object D2 right so you are creating an object D2 it will be set whatever feed 2 you are taking as a input from the user whatever inest 2 you are taking a input from the user so D2 do set values Fe 2 comma in 2 all right so now D1 is invoking ADD D2 so same if I do 98 for D1 and D2 if I have 86 right so D1 the object D1 which is invoking D2 so that means I'm passing 8 and six 8 ft and 6 in is being passed this is what exactly happening over here right so D do in what is D do Ines 6 all right so D do inches will be added with inches what is that inches so that will be paused when the object D1 is invoking ADD so that means the inches is nothing but eight right in the function member function or you call as a method so in the method you call line number 21 you have inches so that eight right so 8 + 6 which is nothing but 14 all right so here you go go you have 14 and feet you are doing feet plus d do feet so your feet is nothing but 9 and D2 do feet is nothing but 8 there it is d. feet so 9 + 8 will be added which is 17 plus you have some more statement result. in /12 so 14/2 is 1 all right so one will be added so here you have 18 and then 14 mod 12 14 mod 12 is 2 so that means 2 18 in 18 ft 2 in all right so this you are displaying so some of the distance result is invoking display and then you have close all right in SC is invoking close so when I give this as a input right 15.0 9.0 8.0 and 7.0 so when I add 9 + 7 is 16 all right 15 + 8 is 23 so 23 + 1 24 this will be 4 16 mod 12 is 4 so 24 ft 4 in I have to get this as the output all right so let us see what you are going to get as output so you can see 24 ft and 4 in right so this we got as an output so the last two classes what we have done how to define a class and how to create an object and we have seen two case studies all right in C++ as well as Java all right so now you can able to write a program so we have not seen the access modifiers like public private soon we are going to see and you have all use the static keyword in the last class right so now you can able to write certain basic programs so take these two as an example and then try to work it out many examples using many many problems right using the concepts whatever you have studied so far right so with this I'm am concluding today's class thank you so much [Music]