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Discussion :: Java.lang Class

  1. What will be the output of the program? 

    try  {     Float f1 = new Float("3.0");     int x = f1.intValue();     byte b = f1.byteValue();     double d = f1.doubleValue();     System.out.println(x + b + d); } catch (NumberFormatException e) /* Line 9 */ {     System.out.println("bad number"); /* Line 11 */ }

    2. A.
    9.0
    B.
    bad number
    C.
    Compilation fails on line 9.
    D.
    Compilation fails on line 11.

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    Workspace

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    Answer : Option A

    Explanation :

    The xxxValue() methods convert any numeric wrapper object's value to any primitive type. When narrowing is necessary, significant bits are dropped and the results are difficult to calculate.

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