[Solved] 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 */ } - Programming - Java Programming - Java.lang Class

Java.lang Class

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Java.lang Class - Finding the output
Java.lang Class - Pointing out the correct statements

Q1: 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 */
}

A 9.0

B bad number

C Compilation fails on line 9.

D Compilation fails on line 11.

ANS:A - 9.0

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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Java.lang Class

Q1: 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 */ }`

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Java.lang Class

Q1: 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 */ }

Solution

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Q1: 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 */ }`

For help Students Orientation
Mcqs Questions