Oracle
1Z0-851 · Question #256
Given: public class Rainbow { public enum MyColor { RED(0xff0000), GREEN(0x00ff00), BLUE(0x0000ff); private final int rgb; MyColor(int rgb) { this.rgb = rgb; } public int getRGB() { return rgb; } }…
The correct answer is B. MyColor treeColor = MyColor.GREEN. See the full explanation below for the reasoning.
Question
Given: public class Rainbow { public enum MyColor { RED(0xff0000), GREEN(0x00ff00), BLUE(0x0000ff); private final int rgb; MyColor(int rgb) { this.rgb = rgb; } public int getRGB() { return rgb; } }; public static void main(String[] args) { // insert code here } } Which code fragment, inserted at line 19, allows the Rainbow class to compile?
Options
- AMyColor skyColor = BLUE;
- BMyColor treeColor = MyColor.GREEN;
- Cif(RED.getRGB() < BLUE.getRGB()) { }
- DCompilation fails due to other error(s) in the code.
- EMyColor purple = new MyColor(0xff00ff);
- FMyColor purple = MyColor.BLUE + MyColor.RED;
How the community answered
(35 responses)- A3% (1)
- B77% (27)
- C11% (4)
- E6% (2)
- F3% (1)
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