About Question enthuware.ocajp.i.v7.2.1377 : (2024)

Note under option #4 says that "the floating point suffices f, F or d, D are used only when using decimal system or hexadecimal and not while using binary".

However, hexadecimal numbers make use of the F character as a digit. How does the compiler know when we meant to use F as a digit and when as a floating point suffix?

Will the number 0x1f be considered integer 31 or float 1.0f?

Integer 31. Compiler has no issues figuring that out because any number starting with 0x is in binary and since there is no 'p' or 'P' in the number, it is a decimal. To write a floating point number is binary, there must be a 'p' or 'P' in the number. Please see this for full details: http://docs.oracle.com/javase/specs/jls ... jls-3.10.2

BTW, 'p' and 'P' are not required for the exam.

HTH,
Paul.

Why are the suffices with binary only invalid for floating points? So, why is this invalid:

float x = 0b100f;

but this valid:

long x = 0b100L;

Also, is there a difference between lower-case and upper-case? Or is f and F the same, d and D the same, l and L the same?

That is how the language designers designed this feature. There is no technical limitation that prevents it.

There is no difference between lower and upper case.

HTH,
Paul.

Hi,

What else should we know about octal, hexadecimal and binary number?

1>I mean should be know the output of the program when they are written?

example :float hexa=0x01dfeacL; System.out.println(hexa);

2>Should we know about the range they take?

example:byte b = 0b1100110011; // Type mismatch: cannot convert from int to byte

Thanks,
GPAR

Maybe the explanation could be clearer on the difference between integer literals in hexadecimal format and floating point literals in hexadecimal format (recognizable by the 'p' or 'P' exponent indicator).

float f1 = 0xffff_ffff; // int literal, assigned to a floatfloat f2 = 0xffff_ffffL; // long literal, assigned to a floatfloat f3 = 0xffff_ffff_ffff_ffffL; // Dittofloat f4 = 0xffff_ffff_ffff_ffff; // Not a valid int literal, because the specified number does not fit in an int!float f5 = 0xffff_ffffp0f; // float literal, assigned to a floatfloat f6 = 0xffff_ffffp0; // double literal, assigned to a float: Possible loss of precision!float f7 = 0xffff_ffff_ffff_ffffp0f; // (very large) float literal, assigned to a float

It is useful to know that hexadecimal integer literals (without the 'p') are interpreted as two's complement, whereas floating point literals are interpreted as unsigned numbers.

System.out.println(f1); // prints -1.0System.out.println(f2); // prints 4.2949673E9System.out.println(f3); // prints -1.0System.out.println(f5); // prints 4.2949673E9System.out.println(f7); // prints 1.8446744E19