Java经典试题

1.public class ReturnIt{returnType methodA(byte x, double y){ //line 2return (short)x/y*2;}}what is valid returnType for methodA in line 2?答案:返回double类型,因为(short)x将byte类型强制转换为short类型,与double类型运算,将会提升为double类型.2.1) class Super{2) public float getNum(){return 3.0f;}3) }4)5) public class Sub extends Super{6)7) }which method, placed at line 6, will cause a compiler error?A. public float getNum(){return 4.0f;}B. public void getNum(){}C. public void getNum(double d){}D. public double getNum(float d){return 4.0d;}Answer:BA属于方法的重写(重写只存在于继承关系中),因为修饰符和参数列表都一样.B出现编译错误,如下:Sub.java:6: Sub 中的getNum() 无法覆盖Super 中的getNum()；正在尝试使用不兼容的返回类型找到：void需要：floatpublic void getNum(){}^1 错误B既不是重写也不是重载,重写需要一样的返回值类型和参数列表,访问修饰符的限制一定要大于被重写方法的访问修饰符（public>protected>default>private）;重载:必须具有不同的参数列表；可以有不同的返回类型，只要参数列表不同就可以了；可以有不同的访问修饰符；把其看做是重载,那么在java中是不能以返回值来区分重载方法的,所以b不对.3.public class IfTest{public static void main(String args[]){int x=3;int y=1;if(x=y)System.out.println("Not equal");elseSystem.out.println("Equal");}}what is the result?Answer:compile error 错误在与if(x=y) 中,应该是x==y; =是赋值符号,==是比较操作符4. public class Foo{public static void main(String args[]){try{return;}finally{ System.out.println("Finally");}}}what is the result?A. print out nothingB. print out "Finally"C. compile errorAnswer:B java的finally块会在return之前执行,无论是否抛出异常且一定执行.5.public class Test{public static String output="";public static void foo(int i){try {if(i==1){throw new Exception();}output +="1";}catch(Exception e){output+="2";return;}finally{output+="3";}output+="4";}public static void main(String args[]){foo(0);foo(1);24)}}what is the value of output at line 24? Answer:13423 如果你想出的答案是134234,那么说明对return的理解有了混淆,return是强制函数返回,本题就是针对foo(),那么当执行到return的话,output+="4"; 就不再执行拉,这个函数就算结束拉.6. public class IfElse{public static void main(String args[]){if(odd(5))System.out.println("odd");elseSystem.out.println("even");}public static int odd(int x){return x%2;}}what is output?Answer:Compile Error7. class ExceptionTest{public static void main(String args[]){try{methodA();}catch(IOException e){System.out.println("caught IOException");}catch(Exception e){System.out.println("caught Exception");}}}If methodA() throws a IOException, what is the result? (其实还应该加上:import java.io.*;) Answer:caught IOException 异常的匹配问题,如果2个catch语句换个位置,那就会报错,catch 只能是越来越大,意思就是说:catch的从上到下的顺序应该是:孙子异常->孩子异常->父亲异常->老祖先异常.这么个顺序.8. int i=1,j=10;do{if(i++>--j) continue;}while(i<5); (注意不要丢了这个分号呦)After Execution, what are the value for i and j?A. i=6 j=5B. i=5 j=5C. i=6 j=4D. i=5 j=6E. i=6 j=6Answer:D9. 1)public class X{2) public Object m(){3) Object o=new Float(3.14F);4) Object[] oa=new Object[1];5) oa[0]=o;6) o=null;7) oa[0]=null;8) System.out.println(oa[0]);9) }10) }which line is the earliest point the object a refered is definitely elibileto be garbage collectioned?A.After line 4B. After line 5C.After line 6D.After line 7E.After line 9(that is,as the method returns)Answer:D如果6) o=null 变成o=9f ,并且把7)去掉,那么8)将会输出什么呢?10. 1) interface Foo{2) int k=0;3) }4) public class Test implements Foo{5) public static void main(String args[]){6) int i;7) Test test = new Test();8) i = test.k;9) i = Test.k;10) i = Foo.k;11) }12) }what is the result? Answer:compile successed and i=0 接口中的int k=0虽然没有访问修饰符,但在接口中默认是static和final的11. what is reserved words in java?A. runB. defaultC. implementD. importAnswer:B,D12. public class Test{public static void main(String[] args){String foo=args[1];Sring bar=args[2];String baz=args[3];}}java Test Red Green Bluewhat is the value of baz?A. baz has value of ""B. baz has value of nullC. baz has value of RedD. baz has value of BlueE. baz has value of GreenF. the code does not compileG. the program throw an exceptionAnswer:G分析:感觉原应该多一些语句吧,至少应该有红绿蓝的赋值语句之类的,才能叫java Test Red Green Blue 才能有后面的选项,所以现在感觉很奇怪,不过就这个样子吧.这个问题在于:数组参数的理解,编译程序没有问题,但是运行这个程序就会出现问题,因为参数args没有给他分配空间那么他的长度应该是0,下面却用拉args[1]........等等的语句,那么定会出现越界错误. 错误如下:Exception in thread "main" ng.ArrayIndexOutOfBoundsException: 1at Test.main(Test.java:4)13. int index=1;int foo[]=new int[3];int bar=foo[index];int baz=bar+index;what is the result?A. baz has a value of 0B. baz has value of 1C. baz has value of 2D. an exception is thrownE. the code will not compileAnswer:B分析:《thinking in java》中的原话：若类的某个成员是基本数据类型，即使没有进行初始化，java也会确保它获得一个默认值，如下表所示：基本类型默认值booleanfalsechar'\u0000'(null)byte(byte)0short(short)0intlong0Lfloat0.0fdouble0.0d千万要小心：当变量作为类的成员使用时，java才确保给定其默认值，。