Plato on Github
Report Home
node_modules/ctype/ctio.js
Maintainability
52.31
Lines of code
1443
Difficulty
296.36
Estimated Errors
8.14
Function weight
By Complexity
By SLOC
/* * rm - Feb 2011 * ctio.js: * * A simple way to read and write simple ctypes. Of course, as you'll find the * code isn't as simple as it might appear. The following types are currently * supported in big and little endian formats: * * uint8_t int8_t * uint16_t int16_t * uint32_t int32_t * float (single precision IEEE 754) * double (double precision IEEE 754) * * This is designed to work in Node and v8. It may in fact work in other * Javascript interpreters (that'd be pretty neat), but it hasn't been tested. * If you find that it does in fact work, that's pretty cool. Try and pass word * back to the original author. * * Note to the reader: If you're tabstop isn't set to 8, parts of this may look * weird. */ /* * Numbers in Javascript have a secret: all numbers must be represented with an * IEEE-754 double. The double has a mantissa with a length of 52 bits with an * implicit one. Thus the range of integers that can be represented is limited * to the size of the mantissa, this makes reading and writing 64-bit integers * difficult, but far from impossible. * * Another side effect of this representation is what happens when you use the * bitwise operators, i.e. shift left, shift right, and, or, etc. In Javascript, * each operand and the result is cast to a signed 32-bit number. However, in * the case of >>> the values are cast to an unsigned number. */ /* * A reminder on endian related issues: * * Big Endian: MSB -> First byte * Little Endian: MSB->Last byte */ var mod_assert = require('assert'); /* * An 8 bit unsigned integer involves doing no significant work. */ function ruint8(buffer, endian, offset) { if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset >= buffer.length) throw (new Error('Trying to read beyond buffer length')); return (buffer[offset]); } /* * For 16 bit unsigned numbers we can do all the casting that we want to do. */ function rgint16(buffer, endian, offset) { var val = 0; if (endian == 'big') { val = buffer[offset] << 8; val |= buffer[offset+1]; } else { val = buffer[offset]; val |= buffer[offset+1] << 8; } return (val); } function ruint16(buffer, endian, offset) { if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 1 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); return (rgint16(buffer, endian, offset)); } /* * Because most bitshifting is done using signed numbers, if we would go into * the realm where we use that 32nd bit, we'll end up going into the negative * range. i.e.: * > 200 << 24 * -939524096 * * Not the value you'd expect. To work around this, we end up having to do some * abuse of the JavaScript standard. in this case, we know that a >>> shift is * defined to cast our value to an *unsigned* 32-bit number. Because of that, we * use that instead to save us some additional math, though it does feel a * little weird and it isn't obvious as to why you woul dwant to do this at * first. */ function rgint32(buffer, endian, offset) { var val = 0; if (endian == 'big') { val = buffer[offset+1] << 16; val |= buffer[offset+2] << 8; val |= buffer[offset+3]; val = val + (buffer[offset] << 24 >>> 0); } else { val = buffer[offset+2] << 16; val |= buffer[offset+1] << 8; val |= buffer[offset]; val = val + (buffer[offset + 3] << 24 >>> 0); } return (val); } function ruint32(buffer, endian, offset) { if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); return (rgint32(buffer, endian, offset)); } /* * Reads a 64-bit unsigned number. The astue observer will note that this * doesn't quite work. Javascript has chosen to only have numbers that can be * represented by a double. A double only has 52 bits of mantissa with an * implicit 1, thus we have up to 53 bits to represent an integer. However, 2^53 * doesn't quite give us what we want. Isn't 53 bits enough for anyone? What * could you have possibly wanted to represent that was larger than that? Oh, * maybe a size? You mean we bypassed the 4 GB limit on file sizes, when did * that happen? * * To get around this egregious language issue, we're going to instead construct * an array of two 32 bit unsigned integers. Where arr[0] << 32 + arr[1] would * give the actual number. However, note that the above code probably won't * produce the desired results because of the way Javascript numbers are * doubles. */ function rgint64(buffer, endian, offset) { var val = new Array(2); if (endian == 'big') { val[0] = ruint32(buffer, endian, offset); val[1] = ruint32(buffer, endian, offset+4); } else { val[0] = ruint32(buffer, endian, offset+4); val[1] = ruint32(buffer, endian, offset); } return (val); } function ruint64(buffer, endian, offset) { if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 7 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); return (rgint64(buffer, endian, offset)); } /* * Signed integer types, yay team! A reminder on how two's complement actually * works. The first bit is the signed bit, i.e. tells us whether or not the * number should be positive or negative. If the two's complement value is * positive, then we're done, as it's equivalent to the unsigned representation. * * Now if the number is positive, you're pretty much done, you can just leverage * the unsigned translations and return those. Unfortunately, negative numbers * aren't quite that straightforward. * * At first glance, one might be inclined to use the traditional formula to * translate binary numbers between the positive and negative values in two's * complement. (Though it doesn't quite work for the most negative value) * Mainly: * - invert all the bits * - add one to the result * * Of course, this doesn't quite work in Javascript. Take for example the value * of -128. This could be represented in 16 bits (big-endian) as 0xff80. But of * course, Javascript will do the following: * * > ~0xff80 * -65409 * * Whoh there, Javascript, that's not quite right. But wait, according to * Javascript that's perfectly correct. When Javascript ends up seeing the * constant 0xff80, it has no notion that it is actually a signed number. It * assumes that we've input the unsigned value 0xff80. Thus, when it does the * binary negation, it casts it into a signed value, (positive 0xff80). Then * when you perform binary negation on that, it turns it into a negative number. * * Instead, we're going to have to use the following general formula, that works * in a rather Javascript friendly way. I'm glad we don't support this kind of * weird numbering scheme in the kernel. * * (BIT-MAX - (unsigned)val + 1) * -1 * * The astute observer, may think that this doesn't make sense for 8-bit numbers * (really it isn't necessary for them). However, when you get 16-bit numbers, * you do. Let's go back to our prior example and see how this will look: * * (0xffff - 0xff80 + 1) * -1 * (0x007f + 1) * -1 * (0x0080) * -1 * * Doing it this way ends up allowing us to treat it appropriately in * Javascript. Sigh, that's really quite ugly for what should just be a few bit * shifts, ~ and &. */ /* * Endianness doesn't matter for 8-bit signed values. We could in fact optimize * this case because the more traditional methods work, but for consistency, * we'll keep doing this the same way. */ function rsint8(buffer, endian, offset) { var neg; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset >= buffer.length) throw (new Error('Trying to read beyond buffer length')); neg = buffer[offset] & 0x80; if (!neg) return (buffer[offset]); return ((0xff - buffer[offset] + 1) * -1); } /* * The 16-bit version requires a bit more effort. In this case, we can leverage * our unsigned code to generate the value we want to return. */ function rsint16(buffer, endian, offset) { var neg, val; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 1 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = rgint16(buffer, endian, offset); neg = val & 0x8000; if (!neg) return (val); return ((0xffff - val + 1) * -1); } /* * We really shouldn't leverage our 32-bit code here and instead utilize the * fact that we know that since these are signed numbers, we can do all the * shifting and binary anding to generate the 32-bit number. But, for * consistency we'll do the same. If we want to do otherwise, we should instead * make the 32 bit unsigned code do the optimization. But as long as there * aren't floats secretly under the hood for that, we /should/ be okay. */ function rsint32(buffer, endian, offset) { var neg, val; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = rgint32(buffer, endian, offset); neg = val & 0x80000000; if (!neg) return (val); return ((0xffffffff - val + 1) * -1); } /* * The signed version of this code suffers from all of the same problems of the * other 64 bit version. */ function rsint64(buffer, endian, offset) { var neg, val; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = rgint64(buffer, endian, offset); neg = val[0] & 0x80000000; if (!neg) return (val); val[0] = (0xffffffff - val[0]) * -1; val[1] = (0xffffffff - val[1] + 1) * -1; /* * If we had the key 0x8000000000000000, that would leave the lower 32 * bits as 0xffffffff, however, since we're goint to add one, that would * actually leave the lower 32-bits as 0x100000000, which would break * our ability to write back a value that we received. To work around * this, if we actually get that value, we're going to bump the upper * portion by 1 and set this to zero. */ mod_assert.ok(val[1] <= 0x100000000); if (val[1] == -0x100000000) { val[1] = 0; val[0]--; } return (val); } /* * We now move onto IEEE 754: The traditional form for floating point numbers * and what is secretly hiding at the heart of everything in this. I really hope * that someone is actually using this, as otherwise, this effort is probably * going to be more wasted. * * One might be tempted to use parseFloat here, but that wouldn't work at all * for several reasons. Mostly due to the way floats actually work, and * parseFloat only actually works in base 10. I don't see base 10 anywhere near * this file. * * In this case we'll implement the single and double precision versions. The * quadruple precision, while probably useful, wouldn't really be accepted by * Javascript, so let's not even waste our time. * * So let's review how this format looks like. A single precision value is 32 * bits and has three parts: * - Sign bit * - Exponent (Using bias notation) * - Mantissa * * |s|eeeeeeee|mmmmmmmmmmmmmmmmmmmmmmmmm| * 31| 30-23 | 22 - 0 | * * The exponent is stored in a biased input. The bias in this case 127. * Therefore, our exponent is equal to the 8-bit value - 127. * * By default, a number is normalized in IEEE, that means that the mantissa has * an implicit one that we don't see. So really the value stored is 1.m. * However, if the exponent is all zeros, then instead we have to shift * everything to the right one and there is no more implicit one. * * Special values: * - Positive Infinity: * Sign: 0 * Exponent: All 1s * Mantissa: 0 * - Negative Infinity: * Sign: 1 * Exponent: All 1s * Mantissa: 0 * - NaN: * Sign: * * Exponent: All 1s * Mantissa: non-zero * - Zero: * Sign: * * Exponent: All 0s * Mantissa: 0 * * In the case of zero, the sign bit determines whether we get a positive or * negative zero. However, since Javascript cannot determine the difference * between the two: i.e. -0 == 0, we just always return 0. * */ function rfloat(buffer, endian, offset) { var bytes = []; var sign, exponent, mantissa, val; var bias = 127; var maxexp = 0xff; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); /* Normalize the bytes to be in endian order */ if (endian == 'big') { bytes[0] = buffer[offset]; bytes[1] = buffer[offset+1]; bytes[2] = buffer[offset+2]; bytes[3] = buffer[offset+3]; } else { bytes[3] = buffer[offset]; bytes[2] = buffer[offset+1]; bytes[1] = buffer[offset+2]; bytes[0] = buffer[offset+3]; } sign = bytes[0] & 0x80; exponent = (bytes[0] & 0x7f) << 1; exponent |= (bytes[1] & 0x80) >>> 7; mantissa = (bytes[1] & 0x7f) << 16; mantissa |= bytes[2] << 8; mantissa |= bytes[3]; /* Check for special cases before we do general parsing */ if (!sign && exponent == maxexp && mantissa === 0) return (Number.POSITIVE_INFINITY); if (sign && exponent == maxexp && mantissa === 0) return (Number.NEGATIVE_INFINITY); if (exponent == maxexp && mantissa !== 0) return (Number.NaN); /* * Javascript really doesn't have support for positive or negative zero. * So we're not going to try and give it to you. That would be just * plain weird. Besides -0 == 0. */ if (exponent === 0 && mantissa === 0) return (0); /* * Now we can deal with the bias and the determine whether the mantissa * has the implicit one or not. */ exponent -= bias; if (exponent == -bias) { exponent++; val = 0; } else { val = 1; } val = (val + mantissa * Math.pow(2, -23)) * Math.pow(2, exponent); if (sign) val *= -1; return (val); } /* * Doubles in IEEE 754 are like their brothers except for a few changes and * increases in size: * - The exponent is now 11 bits * - The mantissa is now 52 bits * - The bias is now 1023 * * |s|eeeeeeeeeee|mmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm| * 63| 62 - 52 | 51 - 0 | * 63| 62 - 52 | 51 - 0 | * * While the size has increased a fair amount, we're going to end up keeping the * same general formula for calculating the final value. As a reminder, this * formula is: * * (-1)^s * (n + m) * 2^(e-b) * * Where: * s is the sign bit * n is (exponent > 0) ? 1 : 0 -- Determines whether we're normalized * or not * m is the mantissa * e is the exponent specified * b is the bias for the exponent * */ function rdouble(buffer, endian, offset) { var bytes = []; var sign, exponent, mantissa, val, lowmant; var bias = 1023; var maxexp = 0x7ff; if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 7 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); /* Normalize the bytes to be in endian order */ if (endian == 'big') { bytes[0] = buffer[offset]; bytes[1] = buffer[offset+1]; bytes[2] = buffer[offset+2]; bytes[3] = buffer[offset+3]; bytes[4] = buffer[offset+4]; bytes[5] = buffer[offset+5]; bytes[6] = buffer[offset+6]; bytes[7] = buffer[offset+7]; } else { bytes[7] = buffer[offset]; bytes[6] = buffer[offset+1]; bytes[5] = buffer[offset+2]; bytes[4] = buffer[offset+3]; bytes[3] = buffer[offset+4]; bytes[2] = buffer[offset+5]; bytes[1] = buffer[offset+6]; bytes[0] = buffer[offset+7]; } /* * We can construct the exponent and mantissa the same way as we did in * the case of a float, just increase the range of the exponent. */ sign = bytes[0] & 0x80; exponent = (bytes[0] & 0x7f) << 4; exponent |= (bytes[1] & 0xf0) >>> 4; /* * This is going to be ugly but then again, we're dealing with IEEE 754. * This could probably be done as a node add on in a few lines of C++, * but oh we'll, we've made it this far so let's be native the rest of * the way... * * What we're going to do is break the mantissa into two parts, the * lower 24 bits and the upper 28 bits. We'll multiply the upper 28 bits * by the appropriate power and then add in the lower 24-bits. Not * really that great. It's pretty much a giant kludge to deal with * Javascript eccentricities around numbers. */ lowmant = bytes[7]; lowmant |= bytes[6] << 8; lowmant |= bytes[5] << 16; mantissa = bytes[4]; mantissa |= bytes[3] << 8; mantissa |= bytes[2] << 16; mantissa |= (bytes[1] & 0x0f) << 24; mantissa *= Math.pow(2, 24); /* Equivalent to << 24, but JS compat */ mantissa += lowmant; /* Check for special cases before we do general parsing */ if (!sign && exponent == maxexp && mantissa === 0) return (Number.POSITIVE_INFINITY); if (sign && exponent == maxexp && mantissa === 0) return (Number.NEGATIVE_INFINITY); if (exponent == maxexp && mantissa !== 0) return (Number.NaN); /* * Javascript really doesn't have support for positive or negative zero. * So we're not going to try and give it to you. That would be just * plain weird. Besides -0 == 0. */ if (exponent === 0 && mantissa === 0) return (0); /* * Now we can deal with the bias and the determine whether the mantissa * has the implicit one or not. */ exponent -= bias; if (exponent == -bias) { exponent++; val = 0; } else { val = 1; } val = (val + mantissa * Math.pow(2, -52)) * Math.pow(2, exponent); if (sign) val *= -1; return (val); } /* * Now that we have gone through the pain of reading the individual types, we're * probably going to want some way to write these back. None of this is going to * be good. But since we have Javascript numbers this should certainly be more * interesting. Though we can constrain this end a little bit more in what is * valid. For now, let's go back to our friends the unsigned value. */ /* * Unsigned numbers seem deceptively easy. Here are the general steps and rules * that we are going to take: * - If the number is negative, throw an Error * - Truncate any floating point portion * - Take the modulus of the number in our base * - Write it out to the buffer in the endian format requested at the offset */ /* * We have to make sure that the value is a valid integer. This means that it is * non-negative. It has no fractional component and that it does not exceed the * maximum allowed value. * * value The number to check for validity * * max The maximum value */ function prepuint(value, max) { if (typeof (value) != 'number') throw (new (Error('cannot write a non-number as a number'))); if (value < 0) throw (new Error('specified a negative value for writing an ' + 'unsigned value')); if (value > max) throw (new Error('value is larger than maximum value for ' + 'type')); if (Math.floor(value) !== value) throw (new Error('value has a fractional component')); return (value); } /* * 8-bit version, classy. We can ignore endianness which is good. */ function wuint8(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepuint(value, 0xff); buffer[offset] = val; } /* * Pretty much the same as the 8-bit version, just this time we need to worry * about endian related issues. */ function wgint16(val, endian, buffer, offset) { if (endian == 'big') { buffer[offset] = (val & 0xff00) >>> 8; buffer[offset+1] = val & 0x00ff; } else { buffer[offset+1] = (val & 0xff00) >>> 8; buffer[offset] = val & 0x00ff; } } function wuint16(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 1 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepuint(value, 0xffff); wgint16(val, endian, buffer, offset); } /* * The 32-bit version is going to have to be a little different unfortunately. * We can't quite bitshift to get the largest byte, because that would end up * getting us caught by the signed values. * * And yes, we do want to subtract out the lower part by default. This means * that when we do the division, it will be treated as a bit shift and we won't * end up generating a floating point value. If we did generate a floating point * value we'd have to truncate it intelligently, this saves us that problem and * may even be somewhat faster under the hood. */ function wgint32(val, endian, buffer, offset) { if (endian == 'big') { buffer[offset] = (val - (val & 0x00ffffff)) / Math.pow(2, 24); buffer[offset+1] = (val >>> 16) & 0xff; buffer[offset+2] = (val >>> 8) & 0xff; buffer[offset+3] = val & 0xff; } else { buffer[offset+3] = (val - (val & 0x00ffffff)) / Math.pow(2, 24); buffer[offset+2] = (val >>> 16) & 0xff; buffer[offset+1] = (val >>> 8) & 0xff; buffer[offset] = val & 0xff; } } function wuint32(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepuint(value, 0xffffffff); wgint32(val, endian, buffer, offset); } /* * Unlike the other versions, we expect the value to be in the form of two * arrays where value[0] << 32 + value[1] would result in the value that we * want. */ function wgint64(value, endian, buffer, offset) { if (endian == 'big') { wgint32(value[0], endian, buffer, offset); wgint32(value[1], endian, buffer, offset+4); } else { wgint32(value[0], endian, buffer, offset+4); wgint32(value[1], endian, buffer, offset); } } function wuint64(value, endian, buffer, offset) { if (value === undefined) throw (new Error('missing value')); if (!(value instanceof Array)) throw (new Error('value must be an array')); if (value.length != 2) throw (new Error('value must be an array of length 2')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 7 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); prepuint(value[0], 0xffffffff); prepuint(value[1], 0xffffffff); wgint64(value, endian, buffer, offset); } /* * We now move onto our friends in the signed number category. Unlike unsigned * numbers, we're going to have to worry a bit more about how we put values into * arrays. Since we are only worrying about signed 32-bit values, we're in * slightly better shape. Unfortunately, we really can't do our favorite binary * & in this system. It really seems to do the wrong thing. For example: * * > -32 & 0xff * 224 * * What's happening above is really: 0xe0 & 0xff = 0xe0. However, the results of * this aren't treated as a signed number. Ultimately a bad thing. * * What we're going to want to do is basically create the unsigned equivalent of * our representation and pass that off to the wuint* functions. To do that * we're going to do the following: * * - if the value is positive * we can pass it directly off to the equivalent wuint * - if the value is negative * we do the following computation: * mb + val + 1, where * mb is the maximum unsigned value in that byte size * val is the Javascript negative integer * * * As a concrete value, take -128. In signed 16 bits this would be 0xff80. If * you do out the computations: * * 0xffff - 128 + 1 * 0xffff - 127 * 0xff80 * * You can then encode this value as the signed version. This is really rather * hacky, but it should work and get the job done which is our goal here. * * Thus the overall flow is: * - Truncate the floating point part of the number * - We don't have to take the modulus, because the unsigned versions will * take care of that for us. And we don't have to worry about that * potentially causing bad things to happen because of sign extension * - Pass it off to the appropriate unsigned version, potentially modifying * the negative portions as necessary. */ /* * A series of checks to make sure we actually have a signed 32-bit number */ function prepsint(value, max, min) { if (typeof (value) != 'number') throw (new (Error('cannot write a non-number as a number'))); if (value > max) throw (new Error('value larger than maximum allowed value')); if (value < min) throw (new Error('value smaller than minimum allowed value')); if (Math.floor(value) !== value) throw (new Error('value has a fractional component')); return (value); } /* * The 8-bit version of the signed value. Overall, fairly straightforward. */ function wsint8(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepsint(value, 0x7f, -0x80); if (val >= 0) wuint8(val, endian, buffer, offset); else wuint8(0xff + val + 1, endian, buffer, offset); } /* * The 16-bit version of the signed value. Also, fairly straightforward. */ function wsint16(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 1 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepsint(value, 0x7fff, -0x8000); if (val >= 0) wgint16(val, endian, buffer, offset); else wgint16(0xffff + val + 1, endian, buffer, offset); } /* * We can do this relatively easily by leveraging the code used for 32-bit * unsigned code. */ function wsint32(value, endian, buffer, offset) { var val; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); val = prepsint(value, 0x7fffffff, -0x80000000); if (val >= 0) wgint32(val, endian, buffer, offset); else wgint32(0xffffffff + val + 1, endian, buffer, offset); } /* * The signed 64 bit integer should by in the same format as when received. * Mainly it should ensure that the value is an array of two integers where * value[0] << 32 + value[1] is the desired number. Furthermore, the two values * need to be equal. */ function wsint64(value, endian, buffer, offset) { var vzpos, vopos; var vals = new Array(2); if (value === undefined) throw (new Error('missing value')); if (!(value instanceof Array)) throw (new Error('value must be an array')); if (value.length != 2) throw (new Error('value must be an array of length 2')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 7 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); /* * We need to make sure that we have the same sign on both values. The * hokiest way to to do this is to multiply the number by +inf. If we do * this, we'll get either +/-inf depending on the sign of the value. * Once we have this, we can compare it to +inf to see if the number is * positive or not. */ vzpos = (value[0] * Number.POSITIVE_INFINITY) == Number.POSITIVE_INFINITY; vopos = (value[1] * Number.POSITIVE_INFINITY) == Number.POSITIVE_INFINITY; /* * If either of these is zero, then we don't actually need this check. */ if (value[0] != 0 && value[1] != 0 && vzpos != vopos) throw (new Error('Both entries in the array must have ' + 'the same sign')); /* * Doing verification for a signed 64-bit integer is actually a big * trickier than it appears. We can't quite use our standard techniques * because we need to compare both sets of values. The first value is * pretty straightforward. If the first value is beond the extremes than * we error out. However, the valid range of the second value varies * based on the first one. If the first value is negative, and *not* the * largest negative value, than it can be any integer within the range [ * 0, 0xffffffff ]. If it is the largest negative number, it must be * zero. * * If the first number is positive, than it doesn't matter what the * value is. We just simply have to make sure we have a valid positive * integer. */ if (vzpos) { prepuint(value[0], 0x7fffffff); prepuint(value[1], 0xffffffff); } else { prepsint(value[0], 0, -0x80000000); prepsint(value[1], 0, -0xffffffff); if (value[0] == -0x80000000 && value[1] != 0) throw (new Error('value smaller than minimum ' + 'allowed value')); } /* Fix negative numbers */ if (value[0] < 0 || value[1] < 0) { vals[0] = 0xffffffff - Math.abs(value[0]); vals[1] = 0x100000000 - Math.abs(value[1]); if (vals[1] == 0x100000000) { vals[1] = 0; vals[0]++; } } else { vals[0] = value[0]; vals[1] = value[1]; } wgint64(vals, endian, buffer, offset); } /* * Now we are moving onto the weirder of these, the float and double. For this * we're going to just have to do something that's pretty weird. First off, we * have no way to get at the underlying float representation, at least not * easily. But that doesn't mean we can't figure it out, we just have to use our * heads. * * One might propose to use Number.toString(2). Of course, this is not really * that good, because the ECMAScript 262 v3 Standard says the following Section * 15.7.4.2-Number.prototype.toString (radix): * * If radix is an integer from 2 to 36, but not 10, the result is a string, the * choice of which is implementation-dependent. * * Well that doesn't really help us one bit now does it? We could use the * standard base 10 version of the string, but that's just going to create more * errors as we end up trying to convert it back to a binary value. So, really * this just means we have to be non-lazy and parse the structure intelligently. * * First off, we can do the basic checks: NaN, positive and negative infinity. * * Now that those are done we can work backwards to generate the mantissa and * exponent. * * The first thing we need to do is determine the sign bit, easy to do, check * whether the value is less than 0. And convert the number to its absolute * value representation. Next, we need to determine if the value is less than * one or greater than or equal to one and from there determine what power was * used to get there. What follows is now specific to floats, though the general * ideas behind this will hold for doubles as well, but the exact numbers * involved will change. * * Once we have that power we can determine the exponent and the mantissa. Call * the value that has the number of bits to reach the power ebits. In the * general case they have the following values: * * exponent 127 + ebits * mantissa value * 2^(23 - ebits) & 0x7fffff * * In the case where the value of ebits is <= -127 we are now in the case where * we no longer have normalized numbers. In this case the values take on the * following values: * * exponent 0 * mantissa value * 2^149 & 0x7fffff * * Once we have the values for the sign, mantissa, and exponent. We reconstruct * the four bytes as follows: * * byte0 sign bit and seven most significant bits from the exp * sign << 7 | (exponent & 0xfe) >>> 1 * * byte1 lsb from the exponent and 7 top bits from the mantissa * (exponent & 0x01) << 7 | (mantissa & 0x7f0000) >>> 16 * * byte2 bits 8-15 (zero indexing) from mantissa * mantissa & 0xff00 >> 8 * * byte3 bits 0-7 from mantissa * mantissa & 0xff * * Once we have this we have to assign them into the buffer in proper endian * order. */ /* * Compute the log base 2 of the value. Now, someone who remembers basic * properties of logarithms will point out that we could use the change of base * formula for logs, and in fact that would be astute, because that's what we'll * do for now. It feels cleaner, albeit it may be less efficient than just * iterating and dividing by 2. We may want to come back and revisit that some * day. */ function log2(value) { return (Math.log(value) / Math.log(2)); } /* * Helper to determine the exponent of the number we're looking at. */ function intexp(value) { return (Math.floor(log2(value))); } /* * Helper to determine the exponent of the fractional part of the value. */ function fracexp(value) { return (Math.floor(log2(value))); } function wfloat(value, endian, buffer, offset) { var sign, exponent, mantissa, ebits; var bytes = []; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 3 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); if (isNaN(value)) { sign = 0; exponent = 0xff; mantissa = 23; } else if (value == Number.POSITIVE_INFINITY) { sign = 0; exponent = 0xff; mantissa = 0; } else if (value == Number.NEGATIVE_INFINITY) { sign = 1; exponent = 0xff; mantissa = 0; } else { /* Well we have some work to do */ /* Thankfully the sign bit is trivial */ if (value < 0) { sign = 1; value = Math.abs(value); } else { sign = 0; } /* Use the correct function to determine number of bits */ if (value < 1) ebits = fracexp(value); else ebits = intexp(value); /* Time to deal with the issues surrounding normalization */ if (ebits <= -127) { exponent = 0; mantissa = (value * Math.pow(2, 149)) & 0x7fffff; } else { exponent = 127 + ebits; mantissa = value * Math.pow(2, 23 - ebits); mantissa &= 0x7fffff; } } bytes[0] = sign << 7 | (exponent & 0xfe) >>> 1; bytes[1] = (exponent & 0x01) << 7 | (mantissa & 0x7f0000) >>> 16; bytes[2] = (mantissa & 0x00ff00) >>> 8; bytes[3] = mantissa & 0x0000ff; if (endian == 'big') { buffer[offset] = bytes[0]; buffer[offset+1] = bytes[1]; buffer[offset+2] = bytes[2]; buffer[offset+3] = bytes[3]; } else { buffer[offset] = bytes[3]; buffer[offset+1] = bytes[2]; buffer[offset+2] = bytes[1]; buffer[offset+3] = bytes[0]; } } /* * Now we move onto doubles. Doubles are similar to floats in pretty much all * ways except that the processing isn't quite as straightforward because we * can't always use shifting, i.e. we have > 32 bit values. * * We're going to proceed in an identical fashion to floats and utilize the same * helper functions. All that really is changing are the specific values that we * use to do the calculations. Thus, to review we have to do the following. * * First get the sign bit and convert the value to its absolute value * representation. Next, we determine the number of bits that we used to get to * the value, branching whether the value is greater than or less than 1. Once * we have that value which we will again call ebits, we have to do the * following in the general case: * * exponent 1023 + ebits * mantissa [value * 2^(52 - ebits)] % 2^52 * * In the case where the value of ebits <= -1023 we no longer use normalized * numbers, thus like with floats we have to do slightly different processing: * * exponent 0 * mantissa [value * 2^1074] % 2^52 * * Once we have determined the sign, exponent and mantissa we can construct the * bytes as follows: * * byte0 sign bit and seven most significant bits form the exp * sign << 7 | (exponent & 0x7f0) >>> 4 * * byte1 Remaining 4 bits from the exponent and the four most * significant bits from the mantissa 48-51 * (exponent & 0x00f) << 4 | mantissa >>> 48 * * byte2 Bits 40-47 from the mantissa * (mantissa >>> 40) & 0xff * * byte3 Bits 32-39 from the mantissa * (mantissa >>> 32) & 0xff * * byte4 Bits 24-31 from the mantissa * (mantissa >>> 24) & 0xff * * byte5 Bits 16-23 from the Mantissa * (mantissa >>> 16) & 0xff * * byte6 Bits 8-15 from the mantissa * (mantissa >>> 8) & 0xff * * byte7 Bits 0-7 from the mantissa * mantissa & 0xff * * Now we can't quite do the right shifting that we want in bytes 1 - 3, because * we'll have extended too far and we'll lose those values when we try and do * the shift. Instead we have to use an alternate approach. To try and stay out * of floating point, what we'll do is say that mantissa -= bytes[4-7] and then * divide by 2^32. Once we've done that we can use binary arithmetic. Oof, * that's ugly, but it seems to avoid using floating point (just based on how v8 * seems to be optimizing for base 2 arithmetic). */ function wdouble(value, endian, buffer, offset) { var sign, exponent, mantissa, ebits; var bytes = []; if (value === undefined) throw (new Error('missing value')); if (endian === undefined) throw (new Error('missing endian')); if (buffer === undefined) throw (new Error('missing buffer')); if (offset === undefined) throw (new Error('missing offset')); if (offset + 7 >= buffer.length) throw (new Error('Trying to read beyond buffer length')); if (isNaN(value)) { sign = 0; exponent = 0x7ff; mantissa = 23; } else if (value == Number.POSITIVE_INFINITY) { sign = 0; exponent = 0x7ff; mantissa = 0; } else if (value == Number.NEGATIVE_INFINITY) { sign = 1; exponent = 0x7ff; mantissa = 0; } else { /* Well we have some work to do */ /* Thankfully the sign bit is trivial */ if (value < 0) { sign = 1; value = Math.abs(value); } else { sign = 0; } /* Use the correct function to determine number of bits */ if (value < 1) ebits = fracexp(value); else ebits = intexp(value); /* * This is a total hack to determine a denormalized value. * Unfortunately, we sometimes do not get a proper value for * ebits, i.e. we lose the values that would get rounded off. * * * The astute observer may wonder why we would be * multiplying by two Math.pows rather than just summing * them. Well, that's to get around a small bug in the * way v8 seems to implement the function. On occasion * doing: * * foo * Math.pow(2, 1023 + 51) * * Causes us to overflow to infinity, where as doing: * * foo * Math.pow(2, 1023) * Math.pow(2, 51) * * Does not cause us to overflow. Go figure. * */ if (value <= 2.225073858507201e-308 || ebits <= -1023) { exponent = 0; mantissa = value * Math.pow(2, 1023) * Math.pow(2, 51); mantissa %= Math.pow(2, 52); } else { /* * We might have gotten fucked by our floating point * logarithm magic. This is rather crappy, but that's * our luck. If we just had a log base 2 or access to * the stupid underlying representation this would have * been much easier and we wouldn't have such stupid * kludges or hacks. */ if (ebits > 1023) ebits = 1023; exponent = 1023 + ebits; mantissa = value * Math.pow(2, -ebits); mantissa *= Math.pow(2, 52); mantissa %= Math.pow(2, 52); } } /* Fill the bytes in backwards to deal with the size issues */ bytes[7] = mantissa & 0xff; bytes[6] = (mantissa >>> 8) & 0xff; bytes[5] = (mantissa >>> 16) & 0xff; mantissa = (mantissa - (mantissa & 0xffffff)) / Math.pow(2, 24); bytes[4] = mantissa & 0xff; bytes[3] = (mantissa >>> 8) & 0xff; bytes[2] = (mantissa >>> 16) & 0xff; bytes[1] = (exponent & 0x00f) << 4 | mantissa >>> 24; bytes[0] = (sign << 7) | (exponent & 0x7f0) >>> 4; if (endian == 'big') { buffer[offset] = bytes[0]; buffer[offset+1] = bytes[1]; buffer[offset+2] = bytes[2]; buffer[offset+3] = bytes[3]; buffer[offset+4] = bytes[4]; buffer[offset+5] = bytes[5]; buffer[offset+6] = bytes[6]; buffer[offset+7] = bytes[7]; } else { buffer[offset+7] = bytes[0]; buffer[offset+6] = bytes[1]; buffer[offset+5] = bytes[2]; buffer[offset+4] = bytes[3]; buffer[offset+3] = bytes[4]; buffer[offset+2] = bytes[5]; buffer[offset+1] = bytes[6]; buffer[offset] = bytes[7]; } } /* * Actually export our work above. One might argue that we shouldn't expose * these interfaces and just force people to use the higher level abstractions * around this work. However, unlike say other libraries we've come across, this * interface has several properties: it makes sense, it's simple, and it's * useful. */ exports.ruint8 = ruint8; exports.ruint16 = ruint16; exports.ruint32 = ruint32; exports.ruint64 = ruint64; exports.wuint8 = wuint8; exports.wuint16 = wuint16; exports.wuint32 = wuint32; exports.wuint64 = wuint64; exports.rsint8 = rsint8; exports.rsint16 = rsint16; exports.rsint32 = rsint32; exports.rsint64 = rsint64; exports.wsint8 = wsint8; exports.wsint16 = wsint16; exports.wsint32 = wsint32; exports.wsint64 = wsint64; exports.rfloat = rfloat; exports.rdouble = rdouble; exports.wfloat = wfloat; exports.wdouble = wdouble;