全國(guó)軟考程序員考試部分例題

1、全國(guó)軟考程序員考試部分例題全國(guó)軟考程序員考試部分例題全國(guó)軟考程序員考試部分例題例題1:choose the three valid ide ntifiers from those listed below.a. idolikethel ongn ameclassb. $bytec. constd. _oke. 3_case解答：a, b, d點(diǎn)評(píng)：java中的標(biāo)示符必須是字母、美元符($)或下劃線(xiàn)(_)開(kāi)頭。關(guān)鍵字與保留字不能作為標(biāo)示符。選項(xiàng)c中的const是java的保留 字，所以不能作標(biāo)示符。選項(xiàng)e中的3_case以數(shù)字開(kāi)頭，違反了 java 的規(guī)則。例題2:how can you for

2、ce garbage collecti on of an object?a. garbage collecti on cannot be forcedb. call system.gc().c. call system.gc(), pass ing in a refere nee to the object tobe garbage collected.d. call run time.gc().e. set all referen ces to the object to new values( n ull, forexample).解答：a點(diǎn)評(píng)：在java中垃圾收集是不能被強(qiáng)迫立即執(zhí)行的。

3、調(diào)用system.gc()或runtime.gc()靜態(tài)方法不能保證垃圾收集器的立即執(zhí) 行，因?yàn)?，也許存在著更高優(yōu)先級(jí)的線(xiàn)程。所以選項(xiàng)b、d不正確。選項(xiàng)c的錯(cuò)誤在于，system.gc()方法是不接受參數(shù)的。選項(xiàng)e中的方法可以使對(duì)象在下次垃圾收集器運(yùn)行時(shí)被收集。例題3:con sider the follow ing class:1. class test(i nt i) 2. void test(i nt i) 3. system.out.pri ntln( i am an in t.);4. 5. void test(stri ng s) 6. system.out.pri ntl n(

4、i am a stri ng.);8.9. public static void mai n(stri ng args) 10. test t=new test();11. char ch= y12. t.test(ch);13. 14. which of the stateme nts below is true?(choose on e.)a. li ne 5 will not compile, because void methods cannot be overridde n.b. li ne 12 will n ot compile, because there is no vers

5、 ion of test() that rakes a char argume nt.c. the code will compile but will throw an excepti on at line 12.d. the code will compile and produce the follow ing output:i am an int.e. the code will compile and produce the follow ing output:i am a stri ng.解答：d點(diǎn)評(píng)：在第12行，16位長(zhǎng)的char型變量ch在編譯時(shí)會(huì)自動(dòng)轉(zhuǎn)化為一個(gè)32位長(zhǎng)的int

6、型，并在運(yùn)行時(shí)傳給void test(int i)方法。例題4:which of the followi ng lines of code will compile withouterror ？a. i nt i=0;if (i) system.out.pr intln(hi )；b.boolea n b=true;boolea n b2二true;if(b=b2) system.out.pr intln( so true ); i=1;int j=2;if(i=1| j=2)system.out.pr in tl n( ok ); i=1;int j=2;if (i=1

7、| j=2)system.out.pr intln( ok );解答：b, c點(diǎn)評(píng)：選項(xiàng)a錯(cuò)，因?yàn)閕f語(yǔ)句后需要一個(gè)boolean類(lèi)型的表達(dá)式。邏輯操作有 tandle extends shape b. public in terface colorable public class shape impleme nts colorable c. public class species public class ani malprivate species species;d. i nteface comp onent class container impleme nts comp onen

8、t private comp onen t childre n;解答：d, e點(diǎn)評(píng)：在java中代碼重用有兩種可能的方式，即組合（has a 關(guān)系）和繼承（is a關(guān)系）。has a關(guān)系是通過(guò)定義類(lèi)的屬性的方 式實(shí)現(xiàn)的；而is a關(guān)系是通過(guò)類(lèi)繼承實(shí)現(xiàn)的。本例中選項(xiàng)a、b、c體現(xiàn)了 is a關(guān)系；選項(xiàng)d、e體現(xiàn)了 has a關(guān)系。例題6:which two stateme nts are true for the class java.util.treeset?（choose two）a. the eleme nts in the collecti on are ordered.b. the c

9、ollectio n is guara nteed to be immutable.c. the elementsin the collectionare guaranteedto beuniq ue.d. the elements unique key.e. the elements synchroni zed解答：a, cin the collectionin the collectionare accessed using aare guaranteedto be點(diǎn)評(píng)：treeset類(lèi)實(shí)現(xiàn)了 set接口。set的特點(diǎn)是其中的元素惟一 選項(xiàng)c正確。由于采用了樹(shù)形存儲(chǔ)方式，將元素有序地組織起

10、來(lái)，所 以選項(xiàng)a也正確。例題7:true or false: readers have methods that can read and retur n floats and doubles.a. tureb. false解答：bfloat點(diǎn)評(píng)：reader/writer 只處理unicode字符的輸入輸出 和double 可以通過(guò)stream 進(jìn)行i/o.what does the follow ingpai nt() method draw?1. public void pain graphics g) 2. g.drawstri ng( any questi on ,10, 0);a.

11、 the string any question? , with its top-left corner at 10,0b. a little squiggle coming down from the top of the comp onent.解答：b點(diǎn)評(píng)：drawstring(string str, int x, int y)方法是使用當(dāng)前的顏色和字符，將str的內(nèi)容顯示出來(lái)，并且最左的字符的基線(xiàn)從(x,y)開(kāi) 始。在本題中，y=0 ,所以基線(xiàn)位于最頂端。我們只能看到下行字母 的一部分，即字母y、q的下半部分。例題9:what happens when you try to compil

12、e and run thefollowi ng applicati on? choose all correct opti ons.1. public class z 2. public static void mai n(stri ng args) 3. new z();4. 5.5. z() 6. z aliasl = this;7. z alias2 = this;8. synchroni zed(aliasl) 9. try 10. alias2.wait();11. system.out.pri ntl n( done wait ing );12. 13. catch (in ter

13、ruptedexcepti on e) 14. system.out.pri ntl n( in terrupted );15. 16. catch (excepti on e) 17. system.out.pri ntln( other excepti on );18. 19. fin ally 20. system.out.pri ntl n(fin ally );21. 22. 23. system.out.pri ntl n( all done );24. 25. a. the applicati on compiles but does n t print anything.b.

14、the applicatio n compiles and print done wait ingc. the applicatio n compiles and print fin allyd. the applicatio n compiles and print all donee. the applicatio n compiles and print in terrupted解答：a點(diǎn)評(píng)：在java中，每一個(gè)對(duì)象都有鎖。任何時(shí)候，該鎖都至多 由一個(gè)線(xiàn)程控制。由于aliasl與alias2指向同一對(duì)象z，在執(zhí)行第 11行前，線(xiàn)程擁有對(duì)象z的鎖。在執(zhí)行完第11行以后，該線(xiàn)程釋放 了對(duì)象z

15、的鎖，進(jìn)入等待池。但此后沒(méi)有線(xiàn)程調(diào)用對(duì)象z的notify。和notifyall()方法，所以該進(jìn)程一直處于等待狀態(tài)，沒(méi)有輸出。例題10 :which statement or statements are true about the code listed below? choose three.1. public class mytextarea exte nds textarea 2. public mytextarea(i nt n rows, int n cols) 3. en ableeve nts(awteve nt.texteve nt_mask);4. 5.5. public

16、 void processtexteve nt(texteve nt te) 6. system.out.pri ntln( process ing a text eve nt.);7. 8. a file calledbe made toa. the source code must appear inmytextarea.javab. betweenlines 2 and 3, a call should super( nrows, n cols) so that the new comp onent will have the correct size.c. at line 6, the retur n type of processtexteve nt() should be declared boolea n, not void.d. betweenlines7 and8, thefollowingcodeshouldappear: retur n true.e. betweenlines7 and8, thefollowingcodeshouldappear: super