atof，是C语言中的一个字符串转化为浮点数的函数，在Java在也有一个对应的实现，就是大家所熟悉的Double.parseDouble(String s)函数。
既然是讲atof的Java实现，肯定脱离不开C语言的实现，引用[我的算法学习之路]中的一句话
“stof是一个简单到爆的“算法””
事实却是如此，字符串转化为浮点数整个的算法核心只有一个，如何将字符‘0‘~‘9‘转化为计算机能识别的数字0~9，而在C语言中有一个很简单的转化方式：int x = (char)c - ‘0‘; 剩下的就是一些异常处理以及如果有效的得到该字符串的位数。
本文核心是想理解Java的实现，借此可以了解Java将字符串转为为浮点数的原理，通过这些代码可以看到，代码编写人注意了每一行代码的变量声明，以及if的逻辑控制提高效率：
<span style="font-size:18px;">public strictfp double doubleValue(){
int kDigits = Math.min( nDigits, maxDecimalDigits+1 );
long lValue;
double dValue;
double rValue, tValue;
// First, check for NaN and Infinity values
if(digits == infinity || digits == notANumber) {
if(digits == notANumber)
return Double.NaN;
else
return (isNegative?Double.NEGATIVE_INFINITY:Double.POSITIVE_INFINITY);
}
else {
if (mustSetRoundDir) {
roundDir = 0;
}
/*
* convert the lead kDigits to a long integer.
*/
// (special performance hack: start to do it using int)
int iValue = (int)digits[0]-(int)'0';
int iDigits = Math.min( kDigits, intDecimalDigits );
for ( int i=1; i < iDigits; i++ ){
iValue = iValue*10 + (int)digits[i]-(int)'0';
}
lValue = (long)iValue;
for ( int i=iDigits; i < kDigits; i++ ){
lValue = lValue*10L + (long)((int)digits[i]-(int)'0');
}
dValue = (double)lValue;
int exp = decExponent-kDigits;
/*
* lValue now contains a long integer with the value of
* the first kDigits digits of the number.
* dValue contains the (double) of the same.
*/
if ( nDigits <= maxDecimalDigits ){
/*
* possibly an easy case.
* We know that the digits can be represented
* exactly. And if the exponent isn't too outrageous,
* the whole thing can be done with one operation,
* thus one rounding error.
* Note that all our constructors trim all leading and
* trailing zeros, so simple values (including zero)
* will always end up here
*/
if (exp == 0 || dValue == 0.0)
return (isNegative)? -dValue : dValue; // small floating integer
else if ( exp >= 0 ){
if ( exp <= maxSmallTen ){
/*
* Can get the answer with one operation,
* thus one roundoff.
*/
rValue = dValue * small10pow[exp];
if ( mustSetRoundDir ){
tValue = rValue / small10pow[exp];
roundDir = ( tValue == dValue ) ? 0
:( tValue < dValue ) ? 1
: -1;
}
return (isNegative)? -rValue : rValue;
}
int slop = maxDecimalDigits - kDigits;
if ( exp <= maxSmallTen+slop ){
/*
* We can multiply dValue by 10^(slop)
* and it is still "small" and exact.
* Then we can multiply by 10^(exp-slop)
* with one rounding.
*/
dValue *= small10pow[slop];
rValue = dValue * small10pow[exp-slop];
if ( mustSetRoundDir ){
tValue = rValue / small10pow[exp-slop];
roundDir = ( tValue == dValue ) ? 0
:( tValue < dValue ) ? 1
: -1;
}
return (isNegative)? -rValue : rValue;
}
/*
* Else we have a hard case with a positive exp.
*/
} else {
if ( exp >= -maxSmallTen ){
/*
* Can get the answer in one division.
*/
rValue = dValue / small10pow[-exp];
tValue = rValue * small10pow[-exp];
if ( mustSetRoundDir ){
roundDir = ( tValue == dValue ) ? 0
:( tValue < dValue ) ? 1
: -1;
}
return (isNegative)? -rValue : rValue;
}
/*
* Else we have a hard case with a negative exp.
*/
}
}
/*
* Harder cases:
* The sum of digits plus exponent is greater than
* what we think we can do with one error.
*
* Start by approximating the right answer by,
* naively, scaling by powers of 10.
*/
if ( exp > 0 ){
if ( decExponent > maxDecimalExponent+1 ){
/*
* Lets face it. This is going to be
* Infinity. Cut to the chase.
*/
return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
}
if ( (exp&15) != 0 ){
dValue *= small10pow[exp&15];
}
if ( (exp>>=4) != 0 ){
int j;
for( j = 0; exp > 1; j++, exp>>=1 ){
if ( (exp&1)!=0)
dValue *= big10pow[j];
}
/*
* The reason for the weird exp > 1 condition
* in the above loop was so that the last multiply
* would get unrolled. We handle it here.
* It could overflow.
*/
double t = dValue * big10pow[j];
if ( Double.isInfinite( t ) ){
/*
* It did overflow.
* Look more closely at the result.
* If the exponent is just one too large,
* then use the maximum finite as our estimate
* value. Else call the result infinity
* and punt it.
* ( I presume this could happen because
* rounding forces the result here to be
* an ULP or two larger than
* Double.MAX_VALUE ).
*/
t = dValue / 2.0;
t *= big10pow[j];
if ( Double.isInfinite( t ) ){
return (isNegative)? Double.NEGATIVE_INFINITY : Double.POSITIVE_INFINITY;
}
t = Double.MAX_VALUE;
}
dValue = t;
}
} else if ( exp < 0 ){
exp = -exp;
if ( decExponent < minDecimalExponent-1 ){
/*
* Lets face it. This is going to be
* zero. Cut to the chase.
*/
return (isNegative)? -0.0 : 0.0;
}
if ( (exp&15) != 0 ){
dValue /= small10pow[exp&15];
}
if ( (exp>>=4) != 0 ){
int j;
for( j = 0; exp > 1; j++, exp>>=1 ){
if ( (exp&1)!=0)
dValue *= tiny10pow[j];
}
/*
* The reason for the weird exp > 1 condition
* in the above loop was so that the last multiply
* would get unrolled. We handle it here.
* It could underflow.
*/
double t = dValue * tiny10pow[j];
if ( t == 0.0 ){
/*
* It did underflow.
* Look more closely at the result.
* If the exponent is just one too small,
* then use the minimum finite as our estimate
* value. Else call the result 0.0
* and punt it.
* ( I presume this could happen because
* rounding forces the result here to be
* an ULP or two less than
* Double.MIN_VALUE ).
*/
t = dValue * 2.0;
t *= tiny10pow[j];
if ( t == 0.0 ){
return (isNegative)? -0.0 : 0.0;
}
t = Double.MIN_VALUE;
}
dValue = t;
}
}
/*
* dValue is now approximately the result.
* The hard part is adjusting it, by comparison
* with FDBigInt arithmetic.
* Formulate the EXACT big-number result as
* bigD0 * 10^exp
*/
FDBigInt bigD0 = new FDBigInt( lValue, digits, kDigits, nDigits );
exp = decExponent - nDigits;
correctionLoop:
while(true){
/* AS A SIDE EFFECT, THIS METHOD WILL SET THE INSTANCE VARIABLES
* bigIntExp and bigIntNBits
*/
FDBigInt bigB = doubleToBigInt( dValue );
/*
* Scale bigD, bigB appropriately for
* big-integer operations.
* Naively, we multiply by powers of ten
* and powers of two. What we actually do
* is keep track of the powers of 5 and
* powers of 2 we would use, then factor out
* common divisors before doing the work.
*/
int B2, B5; // powers of 2, 5 in bigB
int D2, D5; // powers of 2, 5 in bigD
int Ulp2; // powers of 2 in halfUlp.
if ( exp >= 0 ){
B2 = B5 = 0;
D2 = D5 = exp;
} else {
B2 = B5 = -exp;
D2 = D5 = 0;
}
if ( bigIntExp >= 0 ){
B2 += bigIntExp;
} else {
D2 -= bigIntExp;
}
Ulp2 = B2;
// shift bigB and bigD left by a number s. t.
// halfUlp is still an integer.
int hulpbias;
if ( bigIntExp+bigIntNBits <= -expBias+1 ){
// This is going to be a denormalized number
// (if not actually zero).
// half an ULP is at 2^-(expBias+expShift+1)
hulpbias = bigIntExp+ expBias + expShift;
} else {
hulpbias = expShift + 2 - bigIntNBits;
}
B2 += hulpbias;
D2 += hulpbias;
// if there are common factors of 2, we might just as well
// factor them out, as they add nothing useful.
int common2 = Math.min( B2, Math.min( D2, Ulp2 ) );
B2 -= common2;
D2 -= common2;
Ulp2 -= common2;
// do multiplications by powers of 5 and 2
bigB = multPow52( bigB, B5, B2 );
FDBigInt bigD = multPow52( new FDBigInt( bigD0 ), D5, D2 );
//
// to recap:
// bigB is the scaled-big-int version of our floating-point
// candidate.
// bigD is the scaled-big-int version of the exact value
// as we understand it.
// halfUlp is 1/2 an ulp of bigB, except for special cases
// of exact powers of 2
//
// the plan is to compare bigB with bigD, and if the difference
// is less than halfUlp, then we're satisfied. Otherwise,
// use the ratio of difference to halfUlp to calculate a fudge
// factor to add to the floating value, then go 'round again.
//
FDBigInt diff;
int cmpResult;
boolean overvalue;
if ( (cmpResult = bigB.cmp( bigD ) ) > 0 ){
overvalue = true; // our candidate is too big.
diff = bigB.sub( bigD );
if ( (bigIntNBits == 1) && (bigIntExp > -expBias+1) ){
// candidate is a normalized exact power of 2 and
// is too big. We will be subtracting.
// For our purposes, ulp is the ulp of the
// next smaller range.
Ulp2 -= 1;
if ( Ulp2 < 0 ){
// rats. Cannot de-scale ulp this far.
// must scale diff in other direction.
Ulp2 = 0;
diff.lshiftMe( 1 );
}
}
} else if ( cmpResult < 0 ){
overvalue = false; // our candidate is too small.
diff = bigD.sub( bigB );
} else {
// the candidate is exactly right!
// this happens with surprising frequency
break correctionLoop;
}
FDBigInt halfUlp = constructPow52( B5, Ulp2 );
if ( (cmpResult = diff.cmp( halfUlp ) ) < 0 ){
// difference is small.
// this is close enough
if (mustSetRoundDir) {
roundDir = overvalue ? -1 : 1;
}
break correctionLoop;
} else if ( cmpResult == 0 ){
// difference is exactly half an ULP
// round to some other value maybe, then finish
dValue += 0.5*ulp( dValue, overvalue );
// should check for bigIntNBits == 1 here??
if (mustSetRoundDir) {
roundDir = overvalue ? -1 : 1;
}
break correctionLoop;
} else {
// difference is non-trivial.
// could scale addend by ratio of difference to
// halfUlp here, if we bothered to compute that difference.
// Most of the time ( I hope ) it is about 1 anyway.
dValue += ulp( dValue, overvalue );
if ( dValue == 0.0 || dValue == Double.POSITIVE_INFINITY )
break correctionLoop; // oops. Fell off end of range.
continue; // try again.
}
}
return (isNegative)? -dValue : dValue;
}
}</span>
上面就是整个实现的源代码，之所以列出来，省的各位去找了，下面就来一行行解读：
首先跟C语言中的实现一样，去掉了字符串前后的空格：
<span style="font-size:18px;"> in = in.trim()</span>
判断是否为空的字符串：
<span style="font-size:18px;"> int l = in.length();
if (l == 0) throw new NumberFormatException("empty String");</span>
取出第一个字符，判断是否为正负数：
<span style="font-size:18px;"> int i = 0;
switch (c = in.charAt(i)) {
case '-':
isNegative = true;
//FALLTHROUGH
case '+':
i++;
signSeen = true;
}</span>
检查是否为Infinity（无穷大）或者NaN（不明确的数值结果，一般被除数为0会出现这个结果）
<span style="font-size:18px;"> c = in.charAt(i);
if (c == 'N' || c == 'I');</span>
如果既不是Infinity或者NaN，则检查是为十六进制浮点数
<span style="font-size:18px;"> else if (c == '0');</span>
之后就是把字符串中的每个字符拆解出来放到array中，
Java提供了一个方法，将字符array转化为数字，即doubleValue(),
从doubleValue()中可以看到
<span style="font-size:18px;">int iValue = (int)digits[0]-(int)'0';</span>
直接通过了强制类型转换进行数值的转换，剩下的任务就是异常判断以及是否为科学计数法。
总结：从上面可以看出，从算法本身来讲，Java简直弱爆了，一个用C语言十几行代码就能实现的算法，用Java却达到了一百来行的代码。
atof(将字符串转化为浮点数的Java实现)
时间： 2024-10-21 18:25:37