Given:
22. public void go(){
23. String o = "";
24. z:
25. for(int x=0; x<3; x++){
26. for(int y=0; y<2; y++){
27. if(x == 1) break;
28. if(x==2 && y==1) break z;
29. o = o + x + y;
30. }
31. }
32. System.out.println(o);
33. }
What is the result when the go() method is invoked?
(A) 00
(B) 0001
(C) 000120
(D) 00012021
(E) Compilation fails.
F. An exception is thrown at runtime.
答案:登入後查看
統計: A(17), B(39), C(278), D(19), E(48) #1157009
統計: A(17), B(39), C(278), D(19), E(48) #1157009
詳解 (共 2 筆)
#4178497
這題要注意break跳向哪邊
第一次跳的時候回到y第二次跳的是x
跑下來就會變成
X = 0 Y=0
X = 1 跳到 Y的迴圈 Y=0
Y=1 跳到X的迴圈 X =2
Y從0開始跑 Y=0
X<3 所以直接跳出
答案就是 000120
有誤解或有錯請指教
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0
#1263844
Given:
22. public void go(){
23. String o = "";
24. z:
25. for(int x=0; x<3; x++){
26. for(int y=0; y<2; y++){
27. if(x == 1) break;
28. if(x==2 && y==1) break z;
29. o = o + x + y;
30. }
31. }
32. System.out.println(o);
33. }
What is the result when the go() method is invoked?
(A) 00
(B) 0001
(C) 000120
(D) 00012021
(E) Compilation fails. F. An exception is thrown at runtime.
22. public void go(){
23. String o = "";
24. z:
25. for(int x=0; x<3; x++){
26. for(int y=0; y<2; y++){
27. if(x == 1) break;
28. if(x==2 && y==1) break z;
29. o = o + x + y;
30. }
31. }
32. System.out.println(o);
33. }
What is the result when the go() method is invoked?
(A) 00
(B) 0001
(C) 000120
(D) 00012021
(E) Compilation fails. F. An exception is thrown at runtime.
0
0