The conversion of a string to a binary encoded floating point number is not easy, as shown in my last post. The information given there is a bit too much on the theoretical side. I’ll try to change that here with some real code: String to Bigfloat and Bigfloat to String.

The conversion of a string to a float has several stages. The first one is, obviously the parsing of the string. I tried it with JavaScript’s build-in regular expression methods but that made the thing more complicated than necessary. A simple LR parser will do it as well if not even better.

## The String Parser

A floating point number according to IEEE-754 consists of seven parts, some of them are optional.

- The sign of the whole number. One character,
*optional* - The integer part of the number. An arbitrary number of characters,
*optional* - The decimal point (or -mark. In general: radix point). One Character,
*optional* - The fractional part of the number. An arbitrary number of characters,
*optional* - The exponent mark, One character,
*optional* - The sign of the exponent. One character,
*optional* - The exponent part of the number. An arbitrary number of characters,
*optional*

As it is easy to see: anything is optional, so some conditions must exist otherwise even an empty string would have to be considered a proper floating point number.

The conditions:

- Any sign belongs to a number/exponent, only one per number/exponent, only in front of the number/exponent
- An exponent mark belongs to an exponent, only one per exponent
- A decimal point belongs to a number, only one per number, none in the exponent
- A number consists of digits, at least one
- A digit character must not be the same as an exponent mark

Examples:

These are some of correct floating point numbers.

The following numbers are all zero but the sign of the number is relevant and shall

not be dropped. Only minus signs are shown, plus signs are redundant but might exist.

- 0
- 0.
- .0
- 0e0
- 0.e0
- .0e0
- -0
- -0.
- -.0
- -0e0
- -0.e0
- -.0e0
- 0e-0
- 0.e-0
- .0e-0
- -0e-0
- -0.e-0
- -.0e-0
- 0e-100
- 0e100
- -0e-100
- -0e100

Some so called normal numbers

- 123
- 123.
- 1.23
- .123
- 123e3
- 1.23e3
- .123e3
- -123
- -123.
- -1.23
- -.123
- -123e3
- -1.23e3
- -.123e3
- 123e-3
- 123.e-3
- 1.23e-3
- .123e-3
- -123e-3
- -1.23e-3
- -.123e-3

The existence of subnormal/denormal numbers depend on the minimum value of the exponent. There exists such a value in a multi-precision but the exponent can be a 32 bit number or even a big integer of arbitrary size, so subnormal numbers are hard to find in this case.

Not floating pointer numbers.

- .
- 1e
- “”
- .e3
- 1.-e2

Some special values are defined, too.

- Inf
- Infinity
- NaN
- -Inf
- -Infinity
- -NaN

The case of the characters should not matter but it is normally either the above form or all lower-case. The sign *does* matter!

**Note:**One example is the rule-set for the input of `atan2(y,x)`

as defined in ECMAScript 5.1 and, with some slight variations here and there every other implementation of the two valued arcus-tangent function. With some symbols that are *not* a number and for a signed infinity, the two points added to the real line to make it the extended real line and a finite approximation to we get the rule-set from ECMAScript 5.1

.

I hope I did it all correct in my implementation.

### Nasty Typographical Traps

In most prints, the numbers have thousand-separators. Typographically correct is, as far as I remember, one fifths of a unit: 1 000 000. It is rarely seen today, most use some kind of visual mark, mostly a comma when the decimal point is a period and a period when the decimal point is a comma or even something completely different. From that last link to Wikipedia we must accept the fact that the decimal point is not a period everywhere, so we need to localize or restrict it. Localizing is very complicated, but restricting is not as rude as it sounds in the first place. We restrict it to the period (ASCII 0x2e) and think about using the underbar (ASCII 0x5f) as a thousand separator. The problem with the underbar is that e.g.: `_123`

is a proper variable name but this gets caught if we insist on declaring all variables as variables: `let _10 = "ten";`

such that things like `zeta(_10)`

cause a parse error in the line of “Wrong type in argument to zeta()”.

### The Exponent

The character used to mark the beginning of the exponent depends on the base of the number. It is also complicated to do it for all cases, restricting is simpler, as it is always if you are not the one at the front line trying to sell the software you wrote.

We restrict the possible input to the bases 2, (8,) 10, and 16. For the bases 2, 8, and 16 the character “p” as the exponent mark indicates that the exponent itself is encoded in base 10. That makes it an easy choice to restrict the exponent to base 10.

### All Your Base Are Belong to …uhm…sorry, couldn’t resist

A base of the number other then 10 is usually marked with a leading zero followed by a character or—in case of base 8—a digit (which makes base 8 so awkward to implement). The characters are “x” for base 16 and “b” for base 2. We restrict the acceptable bases to 2, (8,) 10 and 16.

Examples:

Base 2: `0b100101.001101p1239`

Base 8: `071234.76125P1239`

Base 10: `97854.7239817e1239`

Base 16: `978aF4.72aF9817p1239`

If it is not already obvious: for such small bases the case of the characters is irrelevant. The encoding of the characters on the other side definitely *is*. We restrict the encoding to ASCII, that is, the byte values 0x2e, 0x30-0x39, 0x41-0x46, 0x50, 0x61-0x66, 0x70, and the underbar at 0x5f.

### Exceptional

The exceptions `Infinity`

, `Inf`

, and `NaN`

together with their signs need some extra handling before the parsing. Well, there is no pressing need to do so but it makes things simpler. So does the extra handling of the base markers at the beginning of the input string and the sign of the number placed at the *very* beginning of the input string.

### The String parser

We have no need for a lot of variables but these few:

// the length of the input string var slen = s.length; // a variable holding the currently read character var c; // a string holding the exponent part of the number var exponent_part = ""; // flag to check if the number has one and only one decimal mark var decimal_point = -1; // digits are digits, so this flag denotes the case that we are // parsing the exponent part of the number var is_expo_part = false; // all digits but the exponent var str_number = ""; // the sign of the number var sign = "+"; // the sign of the exponent var expo_sign = "+"; // the base of the number (base of exponent is always 10) var base = 10; // Index of main loop var i; // Index in string (would be the pointer in C) var k = 0;

We should start with some clean-ups and shortcuts. We do the clean-ups in the main language parser so no clean-ups necessary in the number parser.

The sign, if one exists is always the first symbol, check for it.

// unitary minus/plus comes before base marks (e.g.: "0x") if (s.charAt(k) == "-" || s.charAt(k) == "+") { sign = s.charAt(0); // increase pointer to next index of string k += 1; }

The next shortcut is for the exceptions. Shown for brevity are only two of them. It should even be sufficient, the language parser can do the main work and return either `Inf`

of `NaN`

plus the sign. We do the same here.

if (s.charAt(k) == "I" && s.charAt(k + 1) == "n" && s.charAt(k + 1) == "f") { return sign + "Infinity"; } if (s.charAt(k) == "N" && s.charAt(k + 1) == "a" && s.charAt(k + 1) == "N") { return sign + "NaN"; }

Having that out of the way we can go to the next shortcut: the detection of the base. It could be done with JavaScript’s build-in regular expressions but I wanted it to have as similar to the C implementation as possible.

// check for a leading zero if (s.charAt(k) == "0") { // "x" marks base 16 if (s.charAt(k + 1) == "x") { base = 16; // "0x" are two characters, increment pointer accordingly k += 2; // "b" marks base 2 ("b" as in "binary"?) } else if (s.charAt(k + 1) == "b") { base = 2; // "0b" are two characters, increment pointer accordingly k += 2; } // no leading zeros allowed for octal numbers, except the first // to make things simpler else if (s.charAt(k + 1) && s.charAt(k + 1) != "0" && s.charAt(k + 1) != ".") { base = 8; // "0" is only _one_ character, increment pointer accordingly k += 1; } // it is not a leading zero, it is an actual zero if (s.length == 1) { // return a signed zero (zero is the special number "0e1"") return [base, sign, "0", decimal_point, 1]; } }

Leading zeros are allowed, no problem, just strip them.

while (s.charAt(k) == "0") { k++; } // If all we had were zeros it is an actual zero if (k == slen) { // return a signed zero return [base, sign, "0", decimal_point, 1]; }

We could do another shortcut to handle things like “0.”. “0.0”, or “.0” but that would only clutter the code.

In the hope that we got all special cases treated we can go to the big loop.

One of the problems with JavaScript/ECMAScript is the fact that the strings are encoded in 16 bit Unicode. We trust the main language parser to handle it correctly and to return ASCII only and plainly ignore it.

The main loop starts at the index given by `k`

in the input string and goes up to the end if no fatal error occurred and picks every single character out for treatment. This treatment gets done by a longer `switch`

. An abbreviated loop first to give you a taste of it.

// loop over the string starting at given index for (i = 0; i < slen - k; i++) { // pick a character, could be a 16-bit character-ignored // index offset necessary because we started with i=0 // we also could have started at k, no difference c = s.charAt(i + k); // one large switch to handle all cases switch (c) { case "0": /*...*/ // There are Unicode symbols that look like digits and can be // taken as digits, are even meant to be digits but we insist // on ASCII only here. default: return "Parse error: unknown character = \"" + c + "\""; } }

So it is quite simple, the harder part comes now with the full `switch`

. I use a lot of so called fall-throughs here which could be shortened by a regex but I found out that it is actually shorter this way because you get your parts with very little code with a regular expression but you need more code to handle the individual parts. With such a simple LR-parser you can do it all at once. More or less 😉

for (i = 0; i < slen - k; i++) { c = s.charAt(i + k); switch (c) { case '0': case '1': // These are binary digits, they fit everywhere. // Just check where we are and concat it to the appropriate // variable if (is_expo_part == true) { exponent_part += c; } else { str_number += c; } // nothing else to do-next character, please break; case '2': case '3': case '4': case '5': case '6': case '7': // the base 8 digits that are not fit for // base 2 if (base == 2) { return 'Parse error: digit > base 2'; } // same spiel as above: choose the right part and // concat to variable if (is_expo_part == true) { exponent_part += c; } else { str_number += c; } break; case '8': case '9': // same spiel as above, this time for base 8 and smaller if (base <= 8) { return 'Parse error: digit > base 8'; } if (is_expo_part == true) { exponent_part += c; } else { str_number += c; } break; case 'a': case 'A': case 'b': case 'B': case 'c': case 'C': case 'd': case 'D': // These are hexadecimal characters. // We restricted the exponent to base 10. // The rest is the same as above, this time for // base 10 and smaller if (is_expo_part == true) { return 'Parse error: base in exponent > 10'; } else if (base <= 10) { return 'Parse error: digit > base 10'; } else { str_number += c; } break; case 'e': case 'E': // check if it is a digit or an exponent mark // there can be only one such mark, so raise an error // if it is the second occurrence if (base != 16 && is_expo_part == true) { return 'Parse error: additional exponent part found'; } // We do not have bases larger than 16 so this check // is also valid if (base == 16) { str_number += c; } else { is_expo_part = true; } break; case 'f': case 'F': if (base <= 10) { return 'Parse error: digit > base 10'; } else { str_number += c; } break; case 'p': case 'P': // The exponent mark for bases other than 10 // and an indicator that the exponent is in base 10 if (base == 16) { is_expo_part = true; } else { return 'Parse error: wrong character: "' + c + '"'; } break; case '.': // The decimal point/radix point. // There is only one of it, so raise an error // at the second occurrence if (decimal_point >= 0) { return 'Parse error: additional decimal point found'; } decimal_point = i; break; case '+': case '-': // There is only one sign at the beginning of the exponent part. // It does not check the condition that the sign is at // the beginning of the exponent. Funny thing: it does not matter // here. But the language parser would make an addition out of // it and it would never come down here. if (is_expo_part == true) { expo_sign = c; } else { return 'Parse error: sign at wrong position '; } break; /* case '_': // ignore underbars break; */ default: // Either raise an error or ignore (drop) it. // Raising an error is much cleaner. return 'Parse error: unknown character = "' + c + '"'; } }

We might have nothing at all. Can happen with e.g.: “.” or “.E”, something the main language parser should have caught but did not for some unknown reason.

if (str_number.length == 0) { // No error raised, just return zero instead, for simplicity // Maybe raise a warning, for debugging purposes // console.log("Parse error: number has no digits?"); return [base, sign, "0", decimal_point, 1]; }

The exponent is a small (in the computing sense, meaning that it can be put in a native data type) integer so we can handle it here before it gets forgotten.

if (is_expo_part == true) { exponent_part = parseInt(expo_sign + exponent_part); } else { exponent_part = 0; }

Handle the fraction part. On itself it can have leading zeros e.g.: “0.00000123” but they count, so count them. It can have trailing zeros e.g.: “0.001230000”. They do not count so do not count them.

// we do have a fraction part in the string if (decimal_point >= 0) { // Part the string at the decimal point. // Nearly the same in C where you set two pointers pointing // to the beginning of the individual parts. This is more // complicated in JavaScript, so a substring() will do var parts0 = str_number.substring(0, decimal_point); var parts1 = str_number.substring(decimal_point); k = 0; i = parts1.length - 1; // count leading zeros, integer part while (parts0.charAt(k) == '0') { k++; } // count trailing zeros, fractional part while (parts1.charAt(i) == '0') { i--; } // strip the zeros // As above: substring() instead of pointer juggling parts0 = parts0.substring(k); parts1 = parts1.substring(0, i + 1); k = 0; // count leading zeros, fractional part while (parts1.charAt(k) == '0') { k++; if (k == parts1.length) { break; } } // Move number of trailing zeros to exponent, // that is: 0.0001 = 1e-4 // but only if there is no integer part if (k > 0 && parts0.length == 0) { parts1 = parts1.substring(k); exponent_part -= k; } // Catch a zero here if (parts0.length == 0 && parts1.length - k == 0) { parts0 = '0'; exponent_part = 1; } return [base, sign, parts0, parts1, decimal_point, exponent_part]; } else { k = 0; // Strip leading zeros (not necessary, just in case) while (str_number.charAt(k) == '0') { k++; } str_number = str_number.substring(k); return [base, sign, str_number, '', decimal_point, exponent_part]; }

We should feed this array to a function to get a binary floating point number out of it, otherwise it would be quite a waste, would it not?

This function handles base 10 only for brevity. Other bases are nearly handled the same, some of the work needed for base 10 is not even needed for bases 2, 8, or 16.

var exceptions; // Handle exceptions first // BTW: mixing variable declarations with code // code in JavaScript is considered bad style. // I do not know why, maybe because of the rough grained // variable scopes in JavaScript that might irritate // the beginners? // The number parser returns either an Array or a String. // The latter is a sign to handle an exception if (typeof arrayFromParseNumber === 'string') { // check if it is +-Infinity or +-NaN // using JavaScript regular expression this time for // legibility if (arrayFromParseNumber.search(/[+-]{0,1}Infinity/) >= 0) { exceptions = (new Bigfloat()).setInf(); exceptions.sign = arrayFromParseNumber.match(/{+-}/) [0]; exceptions.sign = (exceptions.sign == '-') ? MP_NEG : MP_ZPOS; exceptions.mantissa.sign = exceptions.sign; return exceptions.sign; } else if (arrayFromParseNumber.search(/[+-]{0,1}NaN/) >= 0) { exceptions = (new Bigfloat()).setNaN(); exceptions.sign = arrayFromParseNumber.match(/[+-]/) [0]; exceptions.sign = (exceptions.sign == '-') ? MP_NEG : MP_ZPOS; exceptions.mantissa.sign = exceptions.sign; return exceptions.sign; } else { // raise some error and... return (new Bigfloat()).setNaN(); } } // Get the content of the function argument var base = arrayFromParseNumber[0]; var sign = arrayFromParseNumber[1]; var int_part = arrayFromParseNumber[2]; var fract_part = arrayFromParseNumber[3]; var decimal_point = arrayFromParseNumber[4]; var exponent = arrayFromParseNumber[5]; // The base var ten = new Bigint(10); var k = 0; var slen = int_part.length + fract_part.length; var numerator, denominator; var num_len, den_len; var s; var fraction; var the_number = new Bigfloat(); // log_2(10), the relation between base 2 and base 10 var log210 = parseFloat('3.321928094887362347870319429489390175865');

Zero is a special number.

// If we got send a zero we are out fast if (int_part == "0" && fract_part == "" && exponent == 1) { the_number.mantissa.sign = (sign == "-") ? MP_NEG : MP_ZPOS; the_number.sign = the_number.mantissa.sign; return the_number; }

We can treat integers and fractions differently.

// E.g. "123e12" but not "1.23e12" despite being an integer, too // The latter gets caught later if (decimal_point < 0 && exponent >= 0) { return handleInteger(exponent); }

Branches for the individual bases. Only base 10 handled here.

if (base == 10) { // Handles base 10, who would have thought. // Scale exponent to be able to treat other integers as integers // 12.34e4 = 1234e2 -> integer // 12.34e2 = 1234e0 -> integer // 12.34e1 = 1234e-1 -> fraction // 12.34e-10 = 1234e-12 -> fraction exponent -= fract_part.length; if (exponent >= 0) { return handleInteger(exponent); } // Convert number string to binary (Bigint), that // will be the numerator // Using the string concatenation of Javascript // for legibility numerator = (int_part + fract_part).toBigint(); // Build denominator as 10 to the power of the decimal // places of the number. // Actually, we could save some money with // 5^x * 2^x = 10^x denominator = ten.pow(Math.abs(exponent)); // Scale it. Both values are already in binary, we can // just take the difference of the bit length. num_len = numerator.highBit(); den_len = denominator.highBit(); // Compute the scale factor s = #(N_r) - #(D_r) s = (num_len - den_len); // Multiply the numerator by 2^(precision - s) where // "precision"" is the working precision of the Bigfloat numerator.lShiftInplace(the_number.precision - s); // Built the Bigfloat's fraction part and round it; // rounding mode is "half to nearest" fraction = roundFraction(numerator, denominator, s); exponent = - (the_number.precision - fraction[1]); // build the Bigfloat and return it. the_number.mantissa = fraction[0]; the_number.mantissa.sign = (sign == '-') ? MP_NEG : MP_ZPOS; the_number.sign = the_number.mantissa.sign; the_number.exponent = exponent; return the_number; } else { // handle bases 2, 8 and 16 // most of the base-10 code can be reused here }

The function `roundFraction(numerator, denominator, s)`

does the real work here but lets take a look at the integer handling first.

// I have put all into one function with sub- and sub-sub-functions // It is possible in JavaScript and sometimes even needed to handle // the scopes. // Just before you wonder where all the variables come from. var handleInteger = function (expo) { var exp; if (arguments.length > 0) { exp = expo; } else { exp = exponent; } // This is redundant code but it is easier to handle // it this way altough it is redundant code. numerator = (int_part + fract_part).toBigint(); numerator = numerator.mul(ten.pow(expo)); // The denominator is one and it needs to be scaled instead // of the numerator denominator = new Bigint(1); // The scale factor s is just the bitlength of the numerator // here s = numerator.highBit(); // Any integers smaller than the work-precsion can be // taken at face value, all others need rounding if (s <= the_number.precision) { numerator.lShiftInplace(the_number.precision - s); exponent = - (the_number.precision - s); the_number.mantissa = numerator; the_number.mantissa.sign = (sign == '-') ? MP_NEG : MP_ZPOS; the_number.sign = the_number.mantissa.sign; the_number.exponent = exponent; return the_number; } else { denominator.lShiftInplace(s - the_number.precision); fraction = roundFraction(numerator, denominator, s); // we can build the exponent from expoent and scale // factor directly exponent = - (the_number.precision - fraction[1]); the_number.mantissa = fraction[0]; the_number.mantissa.sign = (sign == '-') ? MP_NEG : MP_ZPOS; the_number.sign = the_number.mantissa.sign; the_number.exponent = exponent; return the_number; } };

The above function allows for fast treatment of small integers. Where “small” is relative. The minimum precision of the `Bigfloat`

type is 104 that means that 2^104 is the biggest “small” integer handled by this function, a number with 32 decimal digits.

As I mentioned earlier, the rounding function is the heart of it all.

// The function takes the numerator, denominator and scale factor // as arguments in that order var roundFraction = function (num, den, S) { var frac, Q, R, dhalf, cmp; // We might need to do the next lines multiple times // so either loop or use a goto. JavaScript has no goto // so a loop it is. // LOOP: do { // Divide, keep quotient and remainder // numerator = Q * denominator + R frac = num.divrem(den); Q = frac[0]; R = frac[1]; // We need one half of the denominator to compare // the remainder against, so dhalf = denominator/2. // That works only if the denominator is even. There // are some rare cases where that is not the case // It's a cheap test, so just do it. if (den.isEven()) { dhalf = den.rShift(1); } else { dhalf = den.sub(R); } // Rounding. Fixed mode: "half to even" // if(R > dhalf) Q' = Q +1 // if(R == dhalf && Q.isOdd()) Q' = Q +1 cmp = R.cmp(dhalf); if (cmp == MP_GT) { Q.incr(); } if (cmp == MP_EQ) { if (Q.isOdd()) { Q.incr(); } } /* Other rounding modes may need information about the sign, too. */ // Q' might got too large and is > work-precision now // if that is the case set numerator = numerator/2 // and s = s + 1, then goto LOOP if ((Q.highBit() + 1) > the_number.precision) { num.rShiftInplace(1); S++; continue; } // else break out of the loop if one was used instead of // a goto break; } while (true); // return not only the quotient but the scale factor, too return [Q, S]; };

The code calling the functions above

var parsedString = parseNumber(this); var ret = stringToBigfloat(parsedString); // The number is not necessarily normalized // at this point, do it now. ret.normalize(); return ret;

So, all nice and dandy you might say (if you are old enough to know what that term means 😉 ) but what about the other way around? How do I get a nicely (for a wide variety of “nicely”) formatted string out of a `Bigfloat`

?

That is astonishingly easy.

Bigfloat.prototype.toString = function(numbase) { var ret, quot; var log210 = parseFloat("3.321928094887362347870319429489390175865"); var sign = (this.sign == MP_NEG) ? "-" : ""; // getDecimalPrecision() is simply floor(this.precision/log210) var decprec = this.getDecimalPrecision(); var exponent = this.exponent; var signexpo = (this.exponent < 0) ? "-" : ""; var decexpo; var ten = new Bigint(10); var one = new Bigint(1); // Checks for the special value 0e1 if (this.isZero()) { return sign + "0.0E0"; } // We could use the sign of the mantissa directly but that might // not be the correct one. ret = this.mantissa.abs(); // Handle integers that are for sure integers first // What we are doing here is dividing the mantissa--a Bigint--by the // value 2^-exponent. If the exponent is already positive we can // multiply by 2^x which is a simple shift left. if (exponent >= 0) { // lShiftInplace() does nothing if exponent is zero, no check needed. ret.lShiftInplace(exponent); // to avoiud a too complicated formatting at the end if (ret.isZero()) { ret = "00"; } else { ret = ret.toString(); } // The string is alread in base 10, just count the digits decexpo = ret.length - 1; signexpo = ""; } else { // Multiply by 10^(toDec(this.precision)) and divide by 2^exponent // The decimal point is at 10^(toDec(this.precision)). // Don't check for a proper integer; it can't be known if it is a // real one. // What can be done is checking the remainder in the division below-- // if it is zero it is an integer for all intent and purposes decexpo = Math.floor(Math.abs(this.exponent) / log210); // scale ret = ret.mul(ten.pow(decexpo)); // build denominator and divide one.lShiftInplace(Math.abs(exponent)); ret = ret.div(one); // TODO: needs some rounding here, too, or some guard digits in general // NOTE: general guard digits are better--in general. I think. // Dividing by a small integer (smaller than the size of a limb) // gets done in O(n) which is neglible here // Get the last two decimal digits var mod100 = ret.divremInt(100)[1]; // get the last decimal digit for the first branch of rounding var mod10 = mod100 % 10; // round to infinity if (mod10 > 5) { ret = ret.addInt(10 - mod10); } else if (mod10 == 5) { // round half to even, so we need to check // the second to last digit, too mod100 = Math.floor(mod100 / 10) % 10; if (mod100 & 1 == 1) { ret = ret.addInt(10 - mod10); } } // might have been rounded to zero // No, wait, it cannot! Uh, it is cheap, leave it. Just // in case when other rounding modes get implemented. if (ret.isZero()) { ret = "00"; } else { ret = ret.toString(); } // rounding might have added a digit (e.g.: 0.999... <> 1.000...) // so calculate the decimal exponent accordingly if (ret.length > decprec) { decexpo = decprec - decexpo; } else { decexpo = decprec - decexpo - 1; } if (Math.abs(this.exponent) < this.precision) { signexpo = ""; } } if (decexpo == 0) { signexpo = ""; } // I used JavaScript String methods here, but the meaning should // be obvious. ret = sign + ret.slice(0, 1) + "." + ret.slice(1, decprec) + "e" + signexpo + Math.abs(decexpo).toString(); return ret; };

The output format is quite strict, the first digit is always between 1 (one) and 9 (nine) inclusive and all digits show up. E.g.: “0.5” -> “5.000000000000000000000000000000E-1”. But it is a simple string, all information you need are given, so format it as you like.

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