0xp0打印0.0（十六进制浮点数字）

Question

I just wonder why this compiles? and what does it mean since it does compile?

System.out.println(0xp0); // p?

OUTPUT:

0.0

Answer 1

The JLS explains it:

HexadecimalFloatingPointLiteral:
    HexSignificand BinaryExponent FloatTypeSuffixopt

HexSignificand:
    HexNumeral
    HexNumeral .
    0 x HexDigitsopt . HexDigits
    0 X HexDigitsopt . HexDigits

BinaryExponent:
    BinaryExponentIndicator SignedInteger

BinaryExponentIndicator:one of
    p P

Based on the above, I would expect a mandatory .HexDigit before the p , though.

Answer 2

It's a floating point hex literal.

For hexadecimal floating-point literals, at least one digit is required (in either the whole number or the fraction part), and the exponent is mandatory, and the float type suffix is optional. The exponent is indicated by the ASCII letter p or P followed by an optionally signed integer.

See the specification here .

Answer 3

Just for reference, here is how to manually convert the following to a decimal number:

double hfpl = 0x1e.18p4;
System.out.println(hfpl); // 481.5