Lecture 23/2/11

1. Lecture 23/2/11 • Working with an array • Array of user-defined objects (if time permits) 2 MARCH, next Wednesday: Reading week - No lecture 6-7.30, but – room B12 Cruciform is available from 7.30- 9.00: Martin O’Shea will consult over any questions on SP1

2. Advert • In-class open-book Test 9/3/11 subjects • Strings • Methods • Arrays

3. Test 1 results • 39 or less 5 • 40 through 70 21 • 71 or more 9 • Absent 17 • Total 52 Those with marks 50 or less should put more effort: • Look through previous lecture slides before lectures • Try ask questions and do not stop if anything remains unclear; • Visit all lab sessions; • Try do the same tasks by yourself in DCSIS labs;

4. Array Array is an indexed list of elements of the same type; the index is supplied by default (!) A string array nam[ ]: contains both entries and index. String[] nam ={“John”,“Paul”,“George”,“Ringo”};Index:         0          1           2         3 Length (the number of entries) is  4 An integer array age[ ]: int age[ ]= {23, 32, 19, 30, 25, 25, 23, 30};Index:    0  1    2  3   4   5   6  7 Length is  8

5. Work with arrays(1) Data of 5 students: double height[ ]={1.56, 1.72, 1.80, 1.85, 1.90}; //in m double weight[ ]={65.3,80.0,78.1,76.5,112.8}; // in kg Problem: compute the body mass index for all the students, bmi=weight/height2 (in the US, those with bmi between 20 and 25 are considered of normal weight)

6. Work with arrays(2) Loop for is natural with arrays: the index used as the counter double bmi[ ]=new double[5]; for (int I = 0; I < 5; I + +) bmi[I]=weight[I] / (height[I]height[I]); If length of student arrays is not known or is variable, put array’s length whatever it is: double bmi[ ]=new double[height.length]; for (int I = 0; I < height.length; I + +) bmi[I]=weight[I] / (height[I]height[I]);

7. Work with arrays(3) The same result with a method for the bmi: double[ ] bmiind(double h[ ], double w[ ]){ double in[ ]; for (int ii = 0; ii < h.length; ii = ii+1) in[ii]=h[ii]/(w[ii]w[ii]); return in; } Method bmiind is just a frame box; to make it work, one needs to put within a class this: double[ ] bmi=bmiind(weight, height);

8. Work with arrays(4) Still, no result on the screen!!! Printing arrays is not easy in Java: no innate methods for that! double[ ] bmip(double h[ ], double w[ ]){ double in[ ]; for (int a = 0; a < h.length; a++){ in[a]=h[a]/(w[a]w[a]); System.out.println(a + “ ’s bmi is “+in[a]);} return in; } double[ ] b=bmip(weight, height); //Outputs and Prints

9. Finding a maximum in an array   double x[ ]; //assume taken from somewhere int place=-1; //index of the max entry double maxim=-1000; for (int i = 0; i < x.length; i = i+1){ if (x[i] > maxim) {maxim=x[i]; place=i;} } Question: Make it into a method.

10. Finding maximum : a method int place= -1; //for keeping the index of max entry double findMax(double x[ ]){//method’s wrap-up double maxim= -1000; for (int i = 0; i < x.length; i = i+1){ if (x[i] > maxim) {maxim=x[i]; place=i;} } } // Note: a trick with “place”

11. Finding the average in an array   double x[ ]={1.1, 1.2, 1.6, 2.0,1.1}; double average=0; //to accumulate the sum for (int i = 0; i < x.length; i++) average=average+x[i]; //after this, average=7.0; average=average/x.length; //average=7.0/5=1.4; Question: Make it into a method.

12. Finding the average: method   public static void main(String[ ] args){ double x[ ]={1.1, 1.2, 1.6, 2.0,1.1}; double average=MetAv(x);} //average=1.4 public static double MetAv(double[ ] a){ double av=0; for (int i = 0; i < a.length; i++) av=av+a[i]; av=av/a.length; return av; }

13. Constrained average: method   public static void main(String[ ] args){ double x[ ]={1.1, 1.2, 0, 1.6, 2.0, 0}; //av. of no 0s double average=ConAv(x);} public static double ConAv(double[ ] a){ double av=0; int counter=0; for (int i = 0; i < a.length; i++) { if(a[i]!=0){ counter++; av=av+a[i];} } return av/counter; }

14. User defined type reading Follows such classes as: - Oblong and OblongTester, sections 6.3 and 7.2, Charatan and Kans, “Java in two semesters” - Counter and CounterTest, sections 6.3-6.6, Pohl and McDowell, “Java by dissection” - Employee and TwoEmployee, p. 96-112, Ch.3, Farrell, “Java programming”

15. Array of user-defined type(1) • The setting: We have a number of applicants for whom we have separate, but matching, lists of names and id’s organised as arrays. • We would like to develop a new type for an Applicant to hold all individual data of the applicant, in this case just id and name, but it can have as many attributes as it takes. • With this new type, we would like to organise a list of applicants as an array of this type

16. Array of user-defined type(2) • To develop an array of applicants, we need • A named class, say Appl, with variables declared to hold all individual applicant data (A); • A constructor in this class that would take values from the arrays holding id and name information (B); • Representation of the id and name arrays (C); • Generation of an instance of array of new type, Appl (D); • Feeding the Id’s and names into the Appl array (E); • We can show that this does work by printing out data of all the entries in the Appl array.

17. Array of user-defined type(3) • class Appl{ • public int ids; • public String nms; • public Appl(int iden, String nnn){ • ids=iden; • nms=nnn;} • static int[] iden(){ • int ii[]={12, 15, 22, 18, 11}; • return ii;} • static String[] namen(){ • String jj[]={"Aa", "Bb", "Cc", "Dd", "Ee"}; • return jj;}

18. Array of user-defined type(4) • public static void main(String[] args){ • int id[]=iden(); • String name[]=namen(); • Appl[] many=new Appl[id.length]; • for(int i=0;i<id.length;i++) • many[i]=new Appl(id[i],name[i]); • //many – array of Appl type objects. Check: • for(int i=0;i<name.length;i++){ • System.out.println(i +"th applicant data:"); • System.out.println("Id: "+many[i].ids); • System.out.println("Name: "+many[i].nms); • System.out.println(); } • } • }

19. Array of user-defined type(5) Question: • Identify which parts of class Appl correspond • to tasks A, B, C, D and E on slide 16