Dive Into BigDecimal (pending)

nickevin

在Java中处理大数值的精度问题, 可以考虑使用 BigDecimal或BigInteger.
 
环境如下:
OS: Windows XP Service Pack 2
JDK: 1.5.0_10
 
 
BigDecimal bc = new BigDecimal("15323121312.052");BigDecimal bc2 = new BigDecimal("13123234.12");Assert.assertEquals("15336244546.172", bc.add(bc2).toString());Assert.assertEquals("15309998077.932", bc.subtract(bc2).toString());Assert.assertEquals("201088908427219973.61424", bc.multiply(bc2).toString());// java.lang.ArithmeticExceptionAssert.assertEquals("1167.63300661529309057240", bc.divide(bc2).toString());  

运算	结果的首选标度
加	max(addend.scale(), augend.scale())
减	max(minuend.scale(), subtrahend.scale())
乘	multiplier.scale() + multiplicand.scale()
除	dividend.scale() - divisor.scale()

 
从某种意义上说, BigDecimal可以处理任意位数的数值, 查阅Source便可知:
 
BigDecimal实际上内部维护一个数组
/**     * The magnitude of this BigInteger, in <i>big-endian</i> order: the     * zeroth element of this array is the most-significant int of the     * magnitude.  The magnitude must be "minimal" in that the most-significant     * int (<tt>mag[0]</tt>) must be non-zero.  This is necessary to     * ensure that there is exactly one representation for each BigInteger     * value.  Note that this implies that the BigInteger zero has a     * zero-length mag array.     */    int[] mag;  
// 避免小数无限循环, 指定精度位数和RoundingMode bc.divide(bc2, 3, RoundingMode.HALF_UP);