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 *…

Question

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 *…

Accepted Answer

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.

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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Discussion & Comments

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 */ }

More Questions from Java.lang Class

Discussion & Comments

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 */ }