doubleToLongBits(double)

doubleToLongBits(double)

Declaration:
public static long doubleToLongBits(double value)
Description:
Returns a representation of the specified floating-point value according to the IEEE 754 floating-point
“double format” bit layout.
Bit 63 (the bit that is selected by the mask 0x8000000000000000L) represents the sign of the floatingpoint
number. Bits 62-52 (the bits that are selected by the mask 0x7ff0000000000000L) represent the
exponent. Bits 51-0 (the bits that are selected by the mask 0x000fffffffffffffL) represent the
significand (sometimes called the mantissa) of the floating-point number.
If the argument is positive infinity, the result is 0x7ff0000000000000L.
If the argument is negative infinity, the result is 0xfff0000000000000L.
If the argument is NaN, the result is 0x7ff8000000000000L.
In all cases, the result is a long integer that, when given to the longBitsToDouble(long) method,
will produce a floating-point value equal to the argument to doubleToLongBits.
Parameters:
value - a double precision floating-point number.
Returns: the bits that represent the floating-point number.
longBitsToDouble(long)
Declaration:
public static double longBitsToDouble(long bits)
Description:
Returns the double-float corresponding to a given bit representation. The argument is considered to be a
representation of a floating-point value according to the IEEE 754 floating-point “double precision” bit
layout. That floating-point value is returned as the result.
If the argument is 0x7ff0000000000000L, the result is positive infinity.
If the argument is 0xfff0000000000000L, the result is negative infinity.
If the argument is any value in the range 0x7ff0000000000001L through 0x7fffffffffffffffL
or in the range 0xfff0000000000001L through 0xffffffffffffffffL, the result is NaN. All
IEEE 754 NaN values of type double are, in effect, lumped together by the Java programming language
into a single value called NaN.
In all other cases, let s, e, and m be three values that can be computed from the argument:

longBitsToDouble(long)
124
int s = ((bits >> 63) == 0) ? 1 : -1;
int e = (int)((bits >> 52) & 0x7ffL);
long m = (e == 0) ?
(bits & 0xfffffffffffffL) << 1 :
(bits & 0xfffffffffffffL) | 0x10000000000000L;
Then the floating-point result equals the value of the mathematical expression s&#183;m&#183;2e-1075.
Parameters:
bits - any long integer.
Returns: the double floating-point value with the same bit pattern.