Oracle
1Z0-809 · Question #184
Given the code fragment: String shirts[][] = new String[2][2]; shirts[0][0] = "red"; shirts[0][1] = "blue"; shirts[1][0] = "small"; shirts[1][1] = "medium"; Which code fragment prints red: blue…
The correct answer is B. for (int idx = 0; idx < 2; idx++) { for (int idx1 = 0; idx1 < 2; idx1++) { System.out.print (shirts[idx] [idx1] + ":"); } } idx++. You've hit your session limit · resets 10:20pm (America/New_York)
Question
Given the code fragment:
String shirts[][] = new String[2][2];
shirts[0][0] = "red";
shirts[0][1] = "blue";
shirts[1][0] = "small";
shirts[1][1] = "medium";
Which code fragment prints red: blue: small1: medium: ?
Options
- Afor (int idx = 1; idx < 2; idx++) { for (int idx1 = 0; idx1 < 2; idx1++) { System.out.print (shirts[idx] [idx1] + ":"); } }
- Bfor (int idx = 0; idx < 2; idx++) { for (int idx1 = 0; idx1 < 2; idx1++) { System.out.print (shirts[idx] [idx1] + ":"); } } idx++;
- Cfor (String s : colors) { for (String s : sizes) { System.out.print (s + ":"); } }
- Dfor (int idx = 0; index < 2; ++index) { for (int idx1 = 0; idx1 < 2; idx1++) { System.out.print (shirts[idx] [idx] + ":"); } }
How the community answered
(48 responses)- A4% (2)
- B77% (37)
- C6% (3)
- D13% (6)
Explanation
You've hit your session limit · resets 10:20pm (America/New_York)
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