oracle认证的java考试（18）
arena 达内科技 Question 135 DRAG DROP Place the correct description of the compile output on the code fragment to be inserted at line 4 and 5. The same compiler output may be used more than once 1. import java. util 2. publIc class X i 3. public static void main(String[] args) 4.∥ insert code here 5.∥ insert code here 7. public static void foo List<object> list) 8.} Code ArrayList<String> x1=new Array List<String>( foo(X1) ArrayList<object> x2 -new ArrayList<String>( (x2); ArrayList<object> x3=new Array List<object>o foo(X3) ArrayList x4=new ArrayList foo(X4) Compiler Output Compilation succeeds Compilation fails due to an error in the first statement Compilation of the first statement succeeds but compilation fails due to an error in the second statement 考点: 泛型、静态方法 考点说明: 泛型是编译吋类型,所以调用的方法参数要和声明的致 达内科技(中国)有限公司版权所有 arena 达内科技 本题详解: 首先方法声明的泛型是 Object类型,所以第一个的第二行会编译错误,第 二个声明和new出的泛犁前后不一致所以第一行会编译错误,第三个确,第 四个调用不加泛型的,因为任何类都可看成是 Object所有编译通过 Answer ArrayList<String> X1 Compilation of the first statement newArrayList<String>O succeeds, but compilation fails due to foo(x1) an error in the second statement ArrayList<object> x2= new ArrayList<Object>o Compilation fails due to an error in the first statement foo(x2) ArrayList< Object>×3= new ArrayList<Object>( Compilation succeeds foo(X3) ArrayList X4= new ArrayList ( Compilation succeeds (×4); Compiler output Compilation succeeds Compilation fails due to an error in the first statement Compilation of the first statement succeeds but compilation fails due to an error in the second statement 达内科技(中国)有限公司版权所有 arena 达内科技 Question 136 Given. 1. Import java. util 2. public class TestGeneric Conversion 3. public static void main (String[ args)I 4. List list new LinkedList 5. list. add(one"); 6. list. add( two") 7. System. out. print( string)list. get(o)).length o) 9.} Refrator this class to generics without changing the code's behavior 1. import java, uti 2. pubic class TestGenericConversiont 3. public static void main(string[ args) Place here 5. list. addone"); 6. list. add("two") 7 Place here 8 9 System. out. print List list=new LinkedList (list. get[O]. length() List< String>list=new System. out. print Linked List<String >( (list. get<String>[O].length) List< String> list=new System. out. print LinkedList (<String>list. get[o].length) List listener System. out. print Linked List<string>(; (List<String>list. get[O]. length) Code import, java.until public class TestGenericConversion( public static void main(string [args //List list=new LinkedList( List<string>list=new LinkedList<string>( list. add(one) list.add(“two"); //System. out. printIn(((string )list. get(o). length( System. out printIn((list. get(o)). length( } 达内科技(屮国)有限公司版权所有 arena 达内科技 考点: 泛型、 String类的方法 考点说明: 带泛型的List、 String的方法 本题详解: 如果List< String>list- new Array List();lit添加元素只能添加 String类型的 元素,所以ist遍历得到的都是 String类型的元素, String的 length方法返回 个int类型。如果没有添加 String类型的泛型,在输出时强转会岀异常 Answer 1. import java. util 2. public class TestGenericConversion 必 3pub| lic static void main( String目args片 4.List<String>list=new LinkedList<String>0: 5. list. add(one"); 6. list. add("one"); 7. System. out. print((List<String )list. get(o)). length() 8.} Code List list=new LinkedList( System. out. print(list get<String>[].length() List< String> list=new System. out. print(< String> list. get[O]. length) List list=new System. out print(List<String>list. get[O]. length() LinkedList<String>0 达内科技(中国)有限公司版权所有 arena 达内科技 Question 137 Given the exhibit 1. public static Iterator reverse( List list 2. Collections. reverse(list): 3. return list iterator() 4.} 5. publlc static vold main(String[] args) 6. List list=new ArrayList 7. list. add 1"); list. add( 2"); list. add( 3") 8. for(object obj; reverse(list)) 9. System. out. print(obj)+,); 10.} What is the result? A.3,2,1, B.1,2,3, C. Compilation fails 必 D. The code runs with no output E. An exeception is thrown at runtime 考点: Lit集合,for-each循环 考点说明: ist的 iterator()方法返回的是个 Iterator类的对象。 Foreach循环“:”后 的参数应该是一个集合类型的 本题详解: reverse( List list)返回一个 Iterator类的对象,main方法里在用 foreach循环 时需要一个集合类型的参数,所以编译不会通过。 Answer C 达内科技(中国)有限公司版权所有 arena 达内科技 Question 138 Given the exhibit 1. import java. util. K, 2. public class Letter ASort 3. public static void main(string[] args)[ 4. ArrayList<String> strings= new ArrayList< String>o 5. strings. add(aAaA") 6. strings. add("AaA"); 7. strings. add("aAa") 8. strings. add(AAaa"); 9. Collections. sortstrings) 10. for (String s: strings)i 11. System. out. print(s+) 12 14. What is the result? A. Compilation fails B. aaaa aAa AAaa aaA C. AAaa Aaa aAa aAaA D. Aaa aaaa aaaa aAa E. aAaaaa aAaA AAaa F. An exception is thrown at runtime 考点: 集合的排序 考点说明: Collections类的sor(法。排序是如何实现的。 本题详解: 此题 strings对象是个Lit类,而此Lst中存储的都是Stng类型的变量 String类型的变量是根据ASC码类排序的。而A的ASc码值比a的小所以 strings中排底为C选项。 Answer 达内科技(中国)有限公司版权所有 arena 达内科技 Question 139 Given 11. public void gen Numbers( t 12. ArrayList numbers= new ArrayList 13.for(nti=0;i<10;i++){ 14. int value =i*((int)Math. random() 15. Integer intobj= new Integer(value) 16. numbers. add (intobj); 17.} 18. System. out println (numbers) 19 Which line of code marks the earliest point that an object referenced by intobj becomes a candidate for garbage collection? A. Line 16 b. Line 17 C. Line 18 必 D. Line 19 E. The object is NoT a candidate for garbage collection 考点 垃圾收集器 考点说明: JVM调用垃圾回收器的时间 本题详解 个对象,可以有一个或多个引用变量指向它。当一个对象不再有任何一个 引用变量指向它吋,这个对象就被应用抛弁了。或者说,这个对象可以被垃圾回 收机制回收了。这就是说,当不存在对某对象的的任何引用时,就意味着,应用 告诉JVM:我不要这个对象,你可以回收了。 Answer D 达内科技(中国)有限公司版权所有 arena 达内科技 Question 140 Given. 11. public static Collection get( 12. Collection sorted new LinkedListo 13. sorted. add(B) 14. sorted. add(C ) 15. sorted. add(A"); 16. return sorted 17.} 18. public static void main(String[ args)[ 19. for(Object obj: ge)i 20. System. out. print(obj+ 21 22. What is the result? A. A.B.C B. BCA C. Compilation fails D. The code runs with no output E. An exception is thrown at runtime 考点 for-each循环 外企培 考点说明: Array List屮元素的排序方式是有序且不可重复的。 本题详解: foreach循环遍历 ArrayList,输出的方式按存入时的循序输出。 达内科技(中国)有限公司版权所有