BigDecimal 是如何搞定精度缺失的
发布于: 2020 年 10 月 05 日
经常做金额计算的小伙伴,肯定知道计算资金要使用 BigDecimal,而不要使用 double 和 float,因为 double 和 float 会导致小数点后面计算出现精度问题。
为什么会出现精度缺失
Double dou1 = 0.1d;
Double dou2 = 0.2d;
//控制台输出结果(0.30000000000000004)
System.out.println(dou1+dou2);
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表面上我们做的是十进制的加法,实际上计算机在底层把它换算成了二进制,再做运算。但是 0.1 和 0.2 用二进制表示的话位数是无法穷尽的。因此我们看到的 0.1 用二进制表示的某数只是真实的 0.1 的一个近似数。0.2 也是这个道理。所以实际上 0.1+0.2 是两个近似数的相加,因此这个结果也就是 0.3 的近似数啦。这里不做十进制小数转二进制的详细算法,感兴趣的小伙伴可以继续深入研究。
猜测解决思路
既然是因为小数不能精确表示导致的精度缺失,那是不是把小数转化为整数进行计算后,再除以 10 的小数位次方即可解决。
//0.1的10倍转化为1
Double dou1 = 1d;
//0.2的10倍转化为2
Double dou2 = 2d;
//相乘后的结果除以10,控制台输出(0.3),没有出现精度缺失
System.out.println((dou1+dou2)/100);
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看看 BigDecimal 如何处理的
构造器(jdk1.8)
public BigDecimal(char[] in, int offset, int len, MathContext mc) {
// protect against huge length.
if (offset + len > in.length || offset < 0)
throw new NumberFormatException("Bad offset or len arguments for char[] input.");
// This is the primary string to BigDecimal constructor; all
// incoming strings end up here; it uses explicit (inline)
// parsing for speed and generates at most one intermediate
// (temporary) object (a char[] array) for non-compact case.
// Use locals for all fields values until completion
int prec = 0; // record precision value 精度
int scl = 0; // record scale value 小数点后规模
long rs = 0; // the compact value in long
BigInteger rb = null; // the inflated value in BigInteger
// use array bounds checking to handle too-long, len == 0,
// bad offset, etc.
try {
// 支持数字带有符号位,-代表负值 +代表正值
// handle the sign
boolean isneg = false; // assume positive
if (in[offset] == '-') {
isneg = true; // leading minus means negative
offset++;
len--;
} else if (in[offset] == '+') { // leading + allowed
offset++;
len--;
}
// should now be at numeric part of the significand
boolean dot = false; // true when there is a '.'
long exp = 0; // exponent
char c; // current character
//判断数字长度是否超过18位,合法的数字,数字+一个小数点+科学计数法
boolean isCompact = (len <= MAX_COMPACT_DIGITS);
// integer significand array & idx is the index to it. The array
// is ONLY used when we can't use a compact representation.
int idx = 0;
if (isCompact) {
// First compact case, we need not to preserve the character
// and we can just compute the value in place.
for (; len > 0; offset++, len--) {
c = in[offset];
if ((c == '0')) { // have zero
if (prec == 0) //
prec = 1;
else if (rs != 0) {
rs *= 10;
++prec;
} // else digit is a redundant leading zero
if (dot) //如果出现在小数点后,则规模+1
++scl;
} else if ((c >= '1' && c <= '9')) { // have digit
int digit = c - '0';
if (prec != 1 || rs != 0)
++prec; // prec unchanged if preceded by 0s
rs = rs * 10 + digit;
if (dot) //如果出现在小数点后,则规模+1
++scl;
} else if (c == '.') { // have dot
// have dot
if (dot) // two dots
throw new NumberFormatException();
dot = true;
} else if (Character.isDigit(c)) { // slow path
int digit = Character.digit(c, 10);
if (digit == 0) {
if (prec == 0)
prec = 1;
else if (rs != 0) {
rs *= 10;
++prec;
} // else digit is a redundant leading zero
} else {
if (prec != 1 || rs != 0)
++prec; // prec unchanged if preceded by 0s
rs = rs * 10 + digit;
}
if (dot) //如果出现在小数点后,则规模+1
++scl;
} else if ((c == 'e') || (c == 'E')) {
exp = parseExp(in, offset, len);
// Next test is required for backwards compatibility
if ((int) exp != exp) // overflow
throw new NumberFormatException();
break; // [saves a test]
} else {
throw new NumberFormatException();
}
}
if (prec == 0) // no digits found
throw new NumberFormatException();
// Adjust scale if exp is not zero.
if (exp != 0) { // had significant exponent 是否是指数
scl = adjustScale(scl, exp);
}
rs = isneg ? -rs : rs;
int mcp = mc.precision;
int drop = prec - mcp; // prec has range [1, MAX_INT], mcp has range [0, MAX_INT];
// therefore, this subtract cannot overflow
if (mcp > 0 && drop > 0) { // do rounding
while (drop > 0) {
scl = checkScaleNonZero((long) scl - drop);
rs = divideAndRound(rs, LONG_TEN_POWERS_TABLE[drop], mc.roundingMode.oldMode);
prec = longDigitLength(rs);
drop = prec - mcp;
}
}
} else {
char coeff[] = new char[len];
for (; len > 0; offset++, len--) {
c = in[offset];
// have digit
if ((c >= '0' && c <= '9') || Character.isDigit(c)) {
// First compact case, we need not to preserve the character
// and we can just compute the value in place.
if (c == '0' || Character.digit(c, 10) == 0) {
if (prec == 0) {
coeff[idx] = c;
prec = 1;
} else if (idx != 0) {
coeff[idx++] = c;
++prec;
} // else c must be a redundant leading zero
} else {
if (prec != 1 || idx != 0)
++prec; // prec unchanged if preceded by 0s
coeff[idx++] = c;
}
if (dot)
++scl;
continue;
}
// have dot
if (c == '.') {
// have dot
if (dot) // two dots
throw new NumberFormatException();
dot = true;
continue;
}
// exponent expected
if ((c != 'e') && (c != 'E'))
throw new NumberFormatException();
exp = parseExp(in, offset, len);
// Next test is required for backwards compatibility
if ((int) exp != exp) // overflow
throw new NumberFormatException();
break; // [saves a test]
}
// here when no characters left
if (prec == 0) // no digits found
throw new NumberFormatException();
// Adjust scale if exp is not zero.
if (exp != 0) { // had significant exponent
scl = adjustScale(scl, exp);
}
// Remove leading zeros from precision (digits count)
rb = new BigInteger(coeff, isneg ? -1 : 1, prec);
rs = compactValFor(rb);
int mcp = mc.precision;
if (mcp > 0 && (prec > mcp)) {
if (rs == INFLATED) {
int drop = prec - mcp;
while (drop > 0) {
scl = checkScaleNonZero((long) scl - drop);
rb = divideAndRoundByTenPow(rb, drop, mc.roundingMode.oldMode);
rs = compactValFor(rb);
if (rs != INFLATED) {
prec = longDigitLength(rs);
break;
}
prec = bigDigitLength(rb);
drop = prec - mcp;
}
}
if (rs != INFLATED) {
int drop = prec - mcp;
while (drop > 0) {
scl = checkScaleNonZero((long) scl - drop);
rs = divideAndRound(rs, LONG_TEN_POWERS_TABLE[drop], mc.roundingMode.oldMode);
prec = longDigitLength(rs);
drop = prec - mcp;
}
rb = null;
}
}
}
} catch (ArrayIndexOutOfBoundsException e) {
throw new NumberFormatException();
} catch (NegativeArraySizeException e) {
throw new NumberFormatException();
}
this.scale = scl;
this.precision = prec;
this.intCompact = rs;
this.intVal = rb;
}
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计算加法
//判断小数位是否一致
//如果加数小数位小于被加数小数位,则加数*10^差值,然后相加后,取被加数的小数位
//如果加数小数位大于被加数小数位,则被加数*10^差值,然后相加后,取加数的小数位
private static BigDecimal add(final long xs, int scale1, final long ys, int scale2) {
//判断小数位是否一致
long sdiff = (long) scale1 - scale2;
if (sdiff == 0) {
return add(xs, ys, scale1);
} else if (sdiff < 0) {
int raise = checkScale(xs,-sdiff);
long scaledX = longMultiplyPowerTen(xs, raise);
if (scaledX != INFLATED) {
return add(scaledX, ys, scale2);
} else {
BigInteger bigsum = bigMultiplyPowerTen(xs,raise).add(ys);
return ((xs^ys)>=0) ? // same sign test
new BigDecimal(bigsum, INFLATED, scale2, 0)
: valueOf(bigsum, scale2, 0);
}
} else {
int raise = checkScale(ys,sdiff);
long scaledY = longMultiplyPowerTen(ys, raise);
if (scaledY != INFLATED) {
return add(xs, scaledY, scale1);
} else {
BigInteger bigsum = bigMultiplyPowerTen(ys,raise).add(xs);
return ((xs^ys)>=0) ?
new BigDecimal(bigsum, INFLATED, scale1, 0)
: valueOf(bigsum, scale1, 0);
}
}
}
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/**
* Returns a string representation of this {@code BigDecimal}
* without an exponent field. For values with a positive scale,
* the number of digits to the right of the decimal point is used
* to indicate scale. For values with a zero or negative scale,
* the resulting string is generated as if the value were
* converted to a numerically equal value with zero scale and as
* if all the trailing zeros of the zero scale value were present
* in the result.
*
* The entire string is prefixed by a minus sign character '-'
* (<tt>'\u002D'</tt>) if the unscaled value is less than
* zero. No sign character is prefixed if the unscaled value is
* zero or positive.
*
* Note that if the result of this method is passed to the
* {@linkplain #BigDecimal(String) string constructor}, only the
* numerical value of this {@code BigDecimal} will necessarily be
* recovered; the representation of the new {@code BigDecimal}
* may have a different scale. In particular, if this
* {@code BigDecimal} has a negative scale, the string resulting
* from this method will have a scale of zero when processed by
* the string constructor.
*
* (This method behaves analogously to the {@code toString}
* method in 1.4 and earlier releases.)
*
* @return a string representation of this {@code BigDecimal}
* without an exponent field.
* @since 1.5
* @see #toString()
* @see #toEngineeringString()
*/
public String toPlainString() {
if(scale==0) {
if(intCompact!=INFLATED) {
return Long.toString(intCompact);
} else {
return intVal.toString();
}
}
if(this.scale<0) { // No decimal point
if(signum()==0) {
return "0";
}
int tailingZeros = checkScaleNonZero((-(long)scale));
StringBuilder buf;
if(intCompact!=INFLATED) {
buf = new StringBuilder(20+tailingZeros);
buf.append(intCompact);
} else {
String str = intVal.toString();
buf = new StringBuilder(str.length()+tailingZeros);
buf.append(str);
}
for (int i = 0; i < tailingZeros; i++)
buf.append('0');
return buf.toString();
}
String str ;
if(intCompact!=INFLATED) {
str = Long.toString(Math.abs(intCompact));
} else {
str = intVal.abs().toString();
}
return getValueString(signum(), str, scale);
}
/* Returns a digit.digit string */
private String getValueString(int signum, String intString, int scale) {
/* Insert decimal point */
StringBuilder buf;
int insertionPoint = intString.length() - scale;
if (insertionPoint == 0) { /* Point goes right before intVal */
return (signum<0 ? "-0." : "0.") + intString;
} else if (insertionPoint > 0) { /* Point goes inside intVal */
buf = new StringBuilder(intString);
buf.insert(insertionPoint, '.');
if (signum < 0)
buf.insert(0, '-');
} else { /* We must insert zeros between point and intVal */
buf = new StringBuilder(3-insertionPoint + intString.length());
buf.append(signum<0 ? "-0." : "0.");
for (int i=0; i<-insertionPoint; i++)
buf.append('0');
buf.append(intString);
}
return buf.toString();
}
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总结
通过简单分析 BigDecimal 的源码,发现猜测的思路方向是对的,但是 BigDecimal 在异常处理和兼容上考虑的非常完善,支持科学计数法,符号位,哨兵机制防止越界等。
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发布于: 2020 年 10 月 05 日阅读数: 39
版权声明: 本文为 InfoQ 作者【hasWhere】的原创文章。
原文链接:【http://xie.infoq.cn/article/a01fa113557ad82dde3bb6a61】。文章转载请联系作者。
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