2. 交通大學資訊工程學系 Programming in Java More examples, … 蔡文能 交通大學資訊工程學系 tsaiwn@csie.nctu.edu.tw http://www.csie.nctu.edu.tw/~tsaiwn/java//java/

3. 交通大學資訊工程學系 蔡文能 7- 第 2 頁 JavaMore Java Examples Agenda Computing factorial  Loop method  Recursive method Computing primes  Sieve method Sorting an array  Selection Sort  Insertion Sort  Bubble Sort  Quick Sort  Merge Sort

4. 交通大學資訊工程學系 蔡文能 7- 第 3 頁 JavaMore Java Examples Computing Factorial The factorial of an integer is the product of that number and all of the positive integers smaller than it. -0! = 1 -1! = 1... -5! = 5*4*3*2*1 = 120 -6! = 6*5*4*3*2*1 = 720... -50! = 30414093201713378043612608166064768844377641568960512000000000000

5. 交通大學資訊工程學系 蔡文能 7- 第 4 頁 JavaMore Java Examples Factorial -- Iteration public class Factorial { public static long factorial(int n) { long fact = 1; for (int i =2; i <= n; i ++)//for Loop fact *= i;//shorthand for fact=fact*i; return fact; } public class ComputingFactorial { public static void main(String arg[ ]) { long a = Factorial.factorial(Integer.parseInt(arg[0])); System.out.println("Factorial of " + arg[0] + "=" + a); } In separate files

6. 交通大學資訊工程學系 蔡文能 7- 第 5 頁 JavaMore Java Examples Recursion 遞迴概念 很久很久以前, 這裡有一座山, 山裡有一座廟, 廟裡有個老和 尚和小和尚, 小和尚要老和尚說故事給他聽。老和尚就說 : 『 很久很久以前, 這裡有一座山, 山裡友一座廟, 廟裡有個老和 尚和小和尚, 小和尚要老和尚說故事給他聽。..... 莊子與惠子游於濠梁之上。莊子曰：「鯈於出游從容，是 魚之樂也。」 惠子曰 : 『子非魚焉知魚之樂 ? 』 莊子曰 : 『子非我焉知我不知魚之樂 ? 』 惠子曰 : 『子非我焉知我不知你不知魚之樂 ? 』 莊子曰 : 『子非我焉知我不知你不知我知魚之樂 ? 』 惠子曰 : 『子非我焉知我不知你不知我不知你不知魚之樂 ? 』 莊子曰 : 『子非我焉知... !@#$%^&*?...

7. 交通大學資訊工程學系 蔡文能 7- 第 6 頁 JavaMore Java Examples Factorial – Recursion, in Java /** * This class shows a recursive method to compute factorials. This method * calls itself repeatedly based on the formula: n! = n* (n-1)! **/ public class Factorial2 { public static long factorial(int n) { if (n == 1) return 1; else return n*factorial(n - 1); }

8. 交通大學資訊工程學系 蔡文能 7- 第 7 頁 JavaMore Java Examples N Factorial ( Recursive 版 ) long factorial(int n) { if( n < 0) return - factorial( - n); if( n ==1 || n==0) return 1; return n * factorial(n - 1); } /* 告訴我 n-1 階乘, 我就給你 n 階乘 */ #include int main( ){ printf("5! = %ld\n", factorial(5)); } N 階乘就是 N 乘以 N-1 階乘 C 語言版 本

9. 交通大學資訊工程學系 蔡文能 7- 第 8 頁 JavaMore Java Examples GCD -- 歐幾里得的最大公約數 輾轉相除法 (recursive 概念 !) GCD(m, n) = GCD(n, m%n) 但如果 n 是 0 則答案為 m long gcd(long m, long n) { if(n==0) return m; return gcd(n, m%n); } 如何寫成 non-recursive version ?

10. 交通大學資訊工程學系 蔡文能 7- 第 9 頁 JavaMore Java Examples GCD non-recursive version long gcd(long m, long n) { int r; /* remainder */ while(n != 0) { r = m%n; m = n; n = r; } return m; }

11. 交通大學資訊工程學系 蔡文能 7- 第 10 頁 JavaMore Java Examples Recursive 也可以很有趣 (1/3) Fibonacci Series: ( 費氏數列 ) 一開始有一對兔子 小兔子隔一個月可以長大為成兔 每對成兔每隔一個月可生出一對兔子 假設兔子永遠不死, 問第 n 個月時有幾對兔子 ？ 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,. fib(n) 0 1 2 3 4 5 6 7 8 9 10 11 = n

12. 交通大學資訊工程學系 蔡文能 7- 第 11 頁 JavaMore Java Examples Recursive 也可以很有趣 (2/3) long fib(int n) { if( n < 0) return 0; if( n ==1 || n==0) return 1; return fib(n-1)+fib(n-2); } /* 給我 n 我就告訴你第 n 個月時有幾對兔子 */ /* 第 n 個月兔子數 = 第 n-1 個月兔子 + 第 n-2 個月兔子 */ /* 但是, 最開始兩個月例外 ! */

13. 交通大學資訊工程學系 蔡文能 7- 第 12 頁 JavaMore Java Examples Recursive 也可以很有趣 (3/3) Fibonacci Series( 費氏數列 ): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, … 34/55 = 55/89 = 89/144 = 0.618 黃金分割比 :) 神奇的費氏數列 : 任何事物接近這些數字會有變化 請看.. 可 怕 的 巧合 : !? 三 重 魔 力... 民國 34 年台灣光復, 民國 89 年變天 國民黨從日本手上搶回台灣執政剛好 55 年 !

14. 交通大學資訊工程學系 蔡文能 7- 第 13 頁 JavaMore Java Examples 問題與思考 (Recursion) 寫出 non-recursive 版的 Fibonacci function ? 考慮小兔子隔兩個月可以長大為成兔 ? 考慮成兔的懷孕期是兩個月 ? 其他 Recursive 問題  Recursive 版的 binary search  Quick sort  Hanoi tower problem  …

15. 交通大學資訊工程學系 蔡文能 7- 第 14 頁 JavaMore Java Examples Factorial – cache ans in a Table public class Factorial3 { //create an array to cache values 0! Through 20! static long[ ] table = new long[21]; static {ans[0] = 1;} //factorial of 0 is 1 static int last = 0; public static long factorial(int n) { if(n <= last) return ans[n]; long tmp = ans[last]; int k = last; while (k < n) { tmp = (k+1) * tmp; /* (k+1) ! */ if(last < 20) {++last; ans [last ] = tmp;} ++k; } /* while */ return tmp; } Cache 唸作 cash

16. 交通大學資訊工程學系 蔡文能 7- 第 15 頁 JavaMore Java Examples Computing Primes (1/3) Finding the largest prime number smaller than a specified integer: Input integer m, find p  m such that p is a prime and if there is prime p’ > p then p’ must be larger than m. Finding all prime numbers that smaller than a specified integer? 1 2 4 357 68912141516182010 11131719

17. 交通大學資訊工程學系 蔡文能 7- 第 16 頁 JavaMore Java Examples Computing Primes (2/3) Algorithm main idea: find primes by eliminating multiples of the form k  j, where j is a prime smaller than square-root(m) and k is an integer such that k  j  m....... prime 2ij   square-root(m) 2222 3333 4444......... 2222 3333 4444 2222 3333 4444 2 2 2 i i i j j j

18. 交通大學資訊工程學系 蔡文能 7- 第 17 頁 JavaMore Java Examples Computing Primes (3/3) Finding all prime numbers that smaller than a specified integer? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

19. 交通大學資訊工程學系 蔡文能 7- 第 18 頁 JavaMore Java Examples Example: Find Primes (1/2) Import java.lang.*; public class Sieve { public static void main(String[] args) { int max = 100; //Assign a default value try {max = Integer.parseInt(arg[0]);} catch (Exception e) {} //Silently ignore exceptions. //Create an array that specifies whether each number is prime or not. Boolean[] isprime = new boolean[max+1]; //Assume that all numbers are primes, until proven otherwise. for (int i = 0; <= max; i++) isprime[i] = true; //We know that that 0 and 1 are not prime. Make a note of it. isprime[0] = isprime[1] = false;

20. 交通大學資訊工程學系 蔡文能 7- 第 19 頁 JavaMore Java Examples Example: Find Primes (2/2) //To compute all primes less than max, we need to rule out multiples of all //integers less than the square root of max. int n = (int) Math.ceil(Math.sqrt(max)); for (int i = 0; i <= n; i++) { if (isprime[i]) { int k = 2; for (int j = k*i; j <= max; j = (k ++)*i) isprime[j] = false; } } int largest; for (largest = max; !sprime[largest]; largest--); //empty loop body System.out.println("The largest prime less than or equal to " + max + " is " + largest); }

21. 交通大學資訊工程學系 蔡文能 7- 第 20 頁 JavaMore Java Examples Sorting Numbers Sorting: Input n numbers, sort them such that the numbers are ordered increasingly. 3 9 1 6 5 4 8 2 10 7 1 2 3 4 5 6 7 8 9 10

22. 交通大學資訊工程學系 蔡文能 7- 第 21 頁 JavaMore Java Examples Selection Sort (1/4) A simple sorting algorithm Purpose: Input: an array A containing n integers. Output: sorted array. 1.i := 2; 2.Find the least element a from A(i) to A(n); 3.If a is less than A(i - 1), exchange A(i - 1) and a; 4. i := i + 1; if (i <= n) goto step (2).

23. 交通大學資訊工程學系 蔡文能 7- 第 22 頁 JavaMore Java Examples Selection Sort (2/4) 1st step:3 9 1 6 5 4 8 2 10 7 2nd step:1 9 3 6 5 4 8 2 10 7 1 2 3 6 5 4 8 9 10 7 …... swap

24. 交通大學資訊工程學系 蔡文能 7- 第 23 頁 JavaMore Java Examples Selection Sort (3/4) Sorting program: public class SelSort { public static void sort(double[ ] nums) { for(int i = 0; i < nums.length; i++) { int min = i +1; for (int j = i+2; j < nums.length; j++) { if (nums[j] < nums[min]) min = j; } if (nums[i] > nums[min]) { double tmp; tmp = nums[i]; nums[i] = nums[min]; nums[min] = tmp; }

25. 交通大學資訊工程學系 蔡文能 7- 第 24 頁 JavaMore Java Examples Selection Sort (4/4) In separate file: public class SortNumber { public static void main (String[ ] args) { double[ ] nums = new double[10]; //Create an array to hold numbers for(int i = 0; i < nums.length; i++) //Generate random numbers nums[i] = Math.random( )*100; SelSort.sort(nums);//Sort them for (int j = 0; j < nums.length; j++) //Print them out System.out.println(nums [j] ); } Math.random( ) : see next slide

26. 交通大學資訊工程學系 蔡文能 7- 第 25 頁 JavaMore Java Examples Insertion Sort public class InsSort { public static void sort(double[ ] nums) { for(int i = 1; i < nums.length; i++) { double tmp = nums[i]; /* we want to put this in proper position */ k = i – 1; while ((k>=0) && nums[k] > tmp ) { /* smaller than me */ nums[k+1] = nums[k]); /* move it down */ --k ; } nums[k+1] = tmp; /* copy it into correct position */ }

27. 交通大學資訊工程學系 蔡文能 7- 第 26 頁 JavaMore Java Examples Test the Insertion Sort In separate file as previous example (Selection sort): public class SortNumber { public static void main (String[ ] args) { double[ ] nums = new double[10]; //Create an array to hold numbers for(int i = 0; i < nums.length; i++) //Generate random numbers nums[i] = Math.random( )*100; InsSort.sort(nums);//Sort them using insertion sort for (int j = 0; j < nums.length; j++) //Print them out System.out.println(nums [j] ); }

28. 交通大學資訊工程學系 蔡文能 7- 第 27 頁 JavaMore Java Examples Bubble Sort public class BubSort { public static void sort(double[ ] nums) { for(int i = nums.length-2; i >=0; --i ) { int flag = 0; /* Assume no exchange */ for(k=0; k <= i; ++k) { /* walk through all every pass */ if ( nums[k] > nums[k+1] ) { /* incorrect order */ double tmp=nums[k]; nums[k]=nums[k+1]; nums[k+1]=tmp; ++flag; } if(flag == 0) break; /* no need to do next pass */ }

29. 交通大學資訊工程學系 蔡文能 7- 第 28 頁 JavaMore Java Examples Algorithm quick_sort(array A, from, to) Input: from - pointer to the starting position of array A to - pointer to the end position of array A Output: sorted array: A’ 1.Choose any one element as the pivot; 2.Find the first element a = A[i] larger than or equal to pivot from A[from] to A[to]; 3.Find the first element b = A[j] smaller than or equal to pivot from A[to] to A[from]; 4.If i < j then exchange a and b; 5.Repeat step from 2 to 4 until j <= i; 6.If from < j then recursive call quick_sort(A, from, j); 7.If i < to then recursive call quick_sort(A, i, to); Quick Sort (1/6)

30. 交通大學資訊工程學系 蔡文能 7- 第 29 頁 JavaMore Java Examples Quick sort main idea: 1st step:3 1 6 5 4 8 10 7 2nd step: 3 2 1 5 8 9 10 7 3rd step: 3 2 1 4 5 6 8 9 10 7 Choose 5 as pivot from to 29 64 Smaller than any integer right to 5 greater than any integer left to 5 i j Quick Sort (2/6)

31. 交通大學資訊工程學系 蔡文能 7- 第 30 頁 JavaMore Java Examples Quick sort 4th step: 2 4 5 6 109 5th step: 1 2 3 4 5 pivot from to 3 1 87 5 6 7 8 10 9 6th step: pivot from to 7 8 10 9 7th step: 9 10 9 10 8th step: Quick Sort (3/6)

32. 交通大學資訊工程學系 蔡文能 7- 第 31 頁 JavaMore Java Examples public class QuickSorter { public static void sort (int[ ] a, int from, int to) { if ((a == null) || (a.length < 2)) return; int i = from, j = to; int pivot = a[(from + to)/2]; do { while ((i < to) && (a[i] < pivot)) i++; while ((j > from) && (a[j] >= pivot)) j--; if (i < j) { int tmp =a[i]; a [i] = a[j]; a[j] = tmp;} i++; j--; }while (i <= j); if (from < j) sort(a, from, j); if (i < to) sort(a, i, to); } } Quick Sort (4/6)

33. 交通大學資訊工程學系 蔡文能 7- 第 32 頁 JavaMore Java Examples 14 3, 4, 6, 1, 10, 9, 5, 20, 19, 14 1, 12, 2, 15, 21, 13, 18, 17, 8, 16, 3, 4, 6, 1, 10, 9, 5, 8,8, 19, 1, 12, 2, 15, 21, 13, 18, 17,,16, 3, 4, 6, 1, 10, 9, 5, 8, 13,1, 12, 2, 15, 21, 19, 18, 17, 20, 16, i j 3, 4, 6, 1, 10, 9, 5, 8, 13, 1, 12, 2 i j Quick Sort (5/6) 3, 4, 6, 1, 10, 9, 5, 20, 19,14,12, 2, 15, 21, 13, 18, 17, 8, 16, 1 j i 20, 14 i j 3, 4, 6, 1, 10, 9, 5, 8, 13,1, 12, 2, 14, 21, 19, 18, 17, 20, 16, 15 14

34. 交通大學資訊工程學系 蔡文能 7- 第 33 頁 JavaMore Java Examples public class BQSorter { public static void sort (int[ ] a, int from, int to) { if ((a == null) || (a.length = to) return; int k = (from + to)/2; int tmp =a[to]; a [to] = a[k]; a[k] = tmp; int pivot = a[to]; int i = from, j = to-1; while(i < j ) { while ((i < j) && (a[i] < pivot)) i++; while ((i = pivot)) j--; if (i < j) { tmp =a[i]; a [i] = a[j]; a[j] = tmp;} }; tmp =a[i]; a [i] = a[to]; a[to] = tmp; if (from < i-1) sort(a, from, i-1); if (i < to) sort(a, i+1, to); } } Quick Sort (6/6)

35. 交通大學資訊工程學系 蔡文能 7- 第 34 頁 JavaMore Java Examples Merging means the combination of two or more ordered sequence into a single sequence. For example, can merge two sequences: 503, 703, 765 and 087, 512, 677 to obtain a sequence: 087, 503, 512, 677, 703, 765. A simple way to accomplish this is to compare the two smallest items, output the smallest, and then repeat the same process. 503703765 087512677 087 503703765 512677 087503 703765 512677 Merge Sort (1/3)

36. 交通大學資訊工程學系 蔡文能 7- 第 35 頁 JavaMore Java Examples Algorithm Merge(s1, s2) Input: two sequences: s1 - x1  x2...  x m and s2 - y1  y2...  y n Output: a sorted sequence: z1  z2...  z m+n. 1.[initialize]i := 1, j := 1, k := 1; 2.[find smaller]if x i  y j goto step 3, otherwise goto step 5; 3.[output x i ] z k.:= x i, k := k+1, i := i+1. If i  m, goto step 2; 4.[transmit y j ...  y n ] z k,..., z m+n := y j,..., y n. Terminate the algorithm; 5.[output y j ] z k.:= y j, k := k+1, j := j+1. If j  n, goto step 2; 6.[transmit x i ...  x m ] z k,..., z m+n := x i,..., x m. Terminate the algorithm; Merge Sort (2/3)

37. 交通大學資訊工程學系 蔡文能 7- 第 36 頁 JavaMore Java Examples Algorithm Merge-sorting(s) Input: a sequences s = Output: a sorted sequence. 1. If |s| = 1, then return s; 2. k :=  m/2  ; 3. s1 := Merge-sorting(x 1,..., x k ); 4. s2 := Merge-sorting(x k+1,..., x m ); 5. return(Merge(s1, s2)); Merge Sort (3/3)

38. 交通大學資訊工程學系 蔡文能 7- 第 37 頁 JavaMore Java Examples Time complexity of Mergesort Takes roughly n · log 2 n comparisons. Without the shortcut, there is no best or worst case. With the optional shortcut, the best case is when the array is already sorted: takes only (n - 1) comparisons.

39. 交通大學資訊工程學系 蔡文能 7- 第 38 頁 JavaMore Java Examples Binary Search (1/5) Works for an array of numbers or objects that can be compared and are arranged in ascending (or descending) order. A very fast method: only 20 comparisons are needed for an array of 1,000,000 elements; (30 comparisons can handle 1,000,000,000 elements; etc.)

40. 交通大學資訊工程學系 蔡文能 7- 第 39 頁 JavaMore Java Examples Binary Search (2/5) Main idea: “ divide and conquer ” :  compare the target value to the middle element;  if equal, all set  if smaller, apply binary search to the left half;  if larger, apply binary search to the right half.... MI MN MO MS MT NC ND... NJ v NJ

41. 交通大學資訊工程學系 蔡文能 7- 第 40 頁 JavaMore Java Examples Binary Search (3/5) Recursive implementation: public int binarySearch (int arr [ ], int value, int left, int right) { int middle = (left + right) / 2; if (value == arr [middle] ) return middle; else if ( value < arr[middle] && left < middle ) return binarySearch (arr, value, left, middle - 1); else if ( value > arr[middle] && middle < right ) return binarySearch (arr, value, middle + 1, right ); else return -1; // Not found }

42. 交通大學資訊工程學系 蔡文能 7- 第 41 頁 JavaMore Java Examples Binary Search (4/5) Iterative implementation: public int binarySearch (int arr [ ], int value, int left, int right) { while (left <= right) { int middle = (left + right) / 2; if ( value == arr [middle] ) return middle; else if ( value < arr[middle] ) right = middle - 1; else /* if ( value > arr[middle] ) */ left = middle + 1; } return -1; // Not found }

43. 交通大學資訊工程學系 蔡文能 7- 第 42 頁 JavaMore Java Examples Binary Search (5/5) The average number of comparisons is roughly log 2 n The worst-case is log 2 (n + 1) rounded up to an integer (e.g. 2 for n = 3, 3 for n = 7, 4 for n = 15, etc.)

44. 交通大學資訊工程學系 蔡文能 7- 第 43 頁 JavaMore Java Examples PRNG Pseudo Random Number Generator java.lang.Math.random( )  A real (double) number: Uniform random number [0, 1) java.util.Random  A utility class in java.util package  Has many miscellaneous random # generating functions  java.lang.Math.random( ) actually uses nextDouble( ) in this class  nextInt( ) returns a number like int rand( ) in C language  nextGaussian( ) returns a # in standard Normal distribution

45. 交通大學資訊工程學系 蔡文能 7- 第 44 頁 JavaMore Java Examples java.util.Random (1/2) Benchmarks uses the java.util.Random class — a more controlled way to generate random numbers. Constructors: If we set the same “ seed, ” we get the same “ random ” sequence. Random generator = new Random(); Random generator2 = new Random(seed); Default “seed”: System.currentTimeMillis() long seed;

46. 交通大學資訊工程學系 蔡文能 7- 第 45 頁 JavaMore Java Examples java.util.Random (2/2) Methods: int k = generator.nextInt (); int k = generator.nextInt (n); double x = generator.nextDouble (); All 2 32 possible int values are produced with (approximately) equal probability 0  k < n 0  x < 1

47. 交通大學資訊工程學系 蔡文能 7- 第 46 頁 JavaMore Java Examples Normal Distribution Random number

48. 交通大學資訊工程學系 蔡文能 7- 第 47 頁 JavaMore Java Examples java.util.Arrays Provides static methods for dealing with arrays. Works for arrays of numbers, Strings, (and “ comparable ” Objects). Methods: int pos = Arrays.binarySearch (arr, target); Arrays.sort (arr); Arrays.sort (arr, from, to); Arrays.fill (arr, value); // fills arr with a given value Arrays.fill (arr, from, to, value);

49. 交通大學資訊工程學系 蔡文能 7- 第 48 頁 JavaMore Java Examples java.util.List Interface The List library interface describes a list of objects in abstract terms In a list, objects are arranged in sequence obj 0, obj 1,..., obj n - 1 In Java, a list holds references to objects A list can contain duplicate objects (both obj1.equals(obj2) and obj1 == obj2 )

50. 交通大學資訊工程學系 蔡文能 7- 第 49 頁 JavaMore Java Examples List Methods (a Subset) int size(); boolean isEmpty (); boolean add (Object obj); void add (int i, Object obj); Object set(int i, Object obj); Object get(int i); Object remove(int i); boolean contains(Object obj); int indexOf(Object obj); returns true inserts obj as the i-th value; i must be from 0 to size() i must be from 0 to size() - 1 use equals to compare objects

51. 交通大學資訊工程學系 蔡文能 7- 第 50 頁 JavaMore Java Examples java.util.ArrayList (1/6) Implements List using an array Keeps track of the list capacity (the length of the allocated array) and list size (the number of elements currently in the list) "Cat""Hat""Bat" capacity size

52. 交通大學資訊工程學系 蔡文能 7- 第 51 頁 JavaMore Java Examples ArrayList (2/6) Automatically increases (doubles) the capacity when the list runs out of space; allocates a bigger array and copies all the values into it get(i) and set(i, obj) are efficient because an array provides random access to its elements Throws IndexOutOfBoundsException when i < 0 or i  size()

53. 交通大學資訊工程學系 蔡文能 7- 第 52 頁 JavaMore Java Examples ArrayList (3/6) ArrayList holds objects (of any type) If you need to put ints or doubles into a list, use a standard Java array or convert them into Integer or Double objects You have to remember what types of objects your put into the list and may need to cast a retrieved object back into its type

54. 交通大學資訊工程學系 蔡文能 7- 第 53 頁 JavaMore Java Examples ArrayList (4/6) From Java API Docs: JDK1.5: ArrayList(Collection c) Constructs a list containing the elements of the specified collection, in the order they are returned by the collection's iterator.ArrayListCollectionE

55. 交通大學資訊工程學系 蔡文能 7- 第 54 頁 JavaMore Java Examples ArrayList list = new ArrayList (); list.add (new Integer(1)); list.add (new Integer(2)); list.add (new Integer(3)); int sum = 0; for (int i = 0; i < list.size(); i++) sum += ((Integer) list.get(i)). intValue(); ArrayList (5/6) Example 1 Need a cast to Integer in order to call intValue

56. 交通大學資訊工程學系 蔡文能 7- 第 55 頁 JavaMore Java Examples ArrayList words = new ArrayList (4); words.add ("One"); words.add ("Fish"); words.add ("Two"); words.add ("Fish"); int i = 0; while (i < words.size() ) { if ( ”Fish".equals (words.get(i)) ) words.remove(i); else i++; } ArrayList (6/6) Example 2 Shifts all the values after the i-th to the left and decrements the size

57. 交通大學資訊工程學系 蔡文能 7- 第 56 頁 JavaMore Java Examples Generic type

58. 交通大學資訊工程學系 蔡文能 7- 第 57 頁 JavaMore Java Examples Generic Programming Define software components with type parameters  A sorting algorithm has the same structure, regardless of the types being sorted  Stack primitives have the same semantics, regardless of the  objects stored on the stack. Compile time parameters Implementation  Same code used for different parameter values  Different code generated for different values C++ model: template class Java JDK1.5: type inference

59. 交通大學資訊工程學系 蔡文能 7- 第 58 頁 JavaMore Java Examples Exception, More examples, … 謝謝捧場 http://www.csie.nctu.edu.tw/~tsaiwn/java/ 蔡文能