Java Doc for Double.java in  » 6.0-JDK-Modules » j2me » java » lang » Java Source Code / Java DocumentationJava Source Code and Java Documentation

In addition, this class provides several methods for converting a double to a String and a String to a double, as well as other constants and methods useful when dealing with a double.
version:
   12/17/01 (CLDC 1.1)
since:
   JDK1.0, CLDC 1.1

Bit 63 (the bit that is selected by the mask 0x8000000000000000L) represents the sign of the floating-point number.

 Double.longBitsToDouble(0x7fefffffffffffffL)
 

Bit 63 (the bit that is selected by the mask 0x8000000000000000L) represents the sign of the floating-point 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 Double.longBitsToDouble(long) method, will produce a floating-point value equal to the argument to doubleToLongBits.
Parameters:
  value - a double precision floating-point number. the bits that represent the floating-point number.

Note that in most cases, for two instances of class Double, d1 and d2, the value of d1.equals(d2) is true if and only if

 d1.doubleValue() == d2.doubleValue()
 

also has the value true. However, there are two exceptions:

• If d1 and d2 both represent Double.NaN, then the equals method returns true, even though Double.NaN==Double.NaN has the value false.
• If d1 represents +0.0 while d2 represents -0.0, or vice versa, the equals test has the value false, even though +0.0==-0.0 has the value true. This allows hashtables to operate properly.

 (int)(v^(v>>>32))
 

 long v = Double.doubleToLongBits(this.doubleValue());
 

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:

 int s = ((bits >> 63) == 0) ? 1 : -1;
 int e = (int)((bits >> 52) & 0x7ffL);
 long m = (e == 0) ?
 (bits & 0xfffffffffffffL) << 1 :
 (bits & 0xfffffffffffffL) | 0x10000000000000L;
 

• If the argument is NaN, the result is the string "NaN".
• Otherwise, the result is a string that represents the sign and magnitude (absolute value) of the argument. If the sign is negative, the first character of the result is '-' ('\u002d'); if the sign is positive, no sign character appears in the result. As for the magnitude m:
• If m is infinity, it is represented by the characters "Infinity"; thus, positive infinity produces the result "Infinity" and negative infinity produces the result "-Infinity".
• If m is zero, it is represented by the characters "0.0"; thus, negative zero produces the result "-0.0" and positive zero produces the result "0.0".
• If m is greater than or equal to 10^-3 but less than 10⁷, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' (\u002E), followed by one or more decimal digits representing the fractional part of m.
• If m is less than 10^-3 or not less than 10⁷, then it is represented in so-called "computerized scientific notation." Let n be the unique integer such that 10ⁿ<=m<10ⁿ⁺¹; then let a be the mathematically exact quotient of m and 10ⁿ so that 1<=a<10. The magnitude is then represented as the integer part of a, as a single decimal digit, followed by '.' (\u002E), followed by decimal digits representing the fractional part of a, followed by the letter 'E' (\u0045), followed by a representation of n as a decimal integer, as produced by the method Integer.toString(int) .

How many digits must be printed for the fractional part of m or a? There must be at least one digit to represent the fractional part, and beyond that as many, but only as many, more digits as are needed to uniquely distinguish the argument value from adjacent values of type double. That is, suppose that x is the exact mathematical value represented by the decimal representation produced by this method for a finite nonzero argument d. Then d must be the double value nearest to x; or if two double values are equally close to x, then d must be one of them and the least significant bit of the significand of d must be 0.
Parameters:
  d - the double to be converted. a string representation of the argument.

If s is null, then a NullPointerException is thrown.

Leading and trailing whitespace characters in s are ignored. The rest of s should constitute a FloatValue as described by the lexical rule:

 FloatValue:
 Signopt FloatingPointLiteral