转换“浮动"；到没有 Float32Array 的 Javascript 中的字节 [英] Convert "float" to bytes in Javascript without Float32Array

问题描述

好的，我是一个相当烦人的情况，我无法访问诸如 Float32Array 之类的类型化数组，但仍然需要能够将 Javascript 数字转换为字节.现在，我可以很好地处理一个整数，但我不知道如何处理浮点值.

Okay so I'm an a fairly annoying situation where I don't have access to typed arrays such as Float32Array, but still need to be able to convert a Javascript number into bytes. Now, an integer I can handle just fine, but I have no idea how to do it for a floating point value.

我已经解决了相反的问题(将字节转换为浮点数)，但是关于从浮点数转换为字节的文档非常稀缺，因为大多数语言只是让您读取指针或具有用于处理的通用类它.

I've solved the problem of doing it the other way around (bytes into a float), but documentation on converting from float to bytes is pretty scarce, as most language just let you read the pointer or have common classes for handling it.

理想情况下，我希望能够将浮点数转换为 4 字节和 8 字节表示，并选择使用哪一个.但是，可以简单地获取一个数字并将其输出为 8 字节的代码仍然很棒，因为我自己可能会从那里想出 32 位版本.

Ideally I'd like to be able to convert floats into both 4-byte and 8-byte representations, and choose which one to use. However, code that can simply take a number and spit it out as 8-bytes would still be great, as I can probably come up with the 32-bit version myself from there.

推荐答案

好的，所以我真的想通了，所以我将分享我的单精度和双精度解决方案.现在我不能保证它们 100% 符合标准，但它们不需要循环并且似乎工作得很好:

Okay, so I actually figured it out, so I'll share my solution for single and double precision. Now I can't guarantee that they're 100% standards compliant, but they require no loops and seem to work just fine:

单精度(给定一个十进制值输出一个具有二进制表示的单个 32 位大端整数):

Single precision (given a decimal value outputs a single 32-bit big endian integer with the binary representation):

function toFloat32(value) {
    var bytes = 0;
    switch (value) {
        case Number.POSITIVE_INFINITY: bytes = 0x7F800000; break;
        case Number.NEGATIVE_INFINITY: bytes = 0xFF800000; break;
        case +0.0: bytes = 0x40000000; break;
        case -0.0: bytes = 0xC0000000; break;
        default:
            if (Number.isNaN(value)) { bytes = 0x7FC00000; break; }

            if (value <= -0.0) {
                bytes = 0x80000000;
                value = -value;
            }

            var exponent = Math.floor(Math.log(value) / Math.log(2));
            var significand = ((value / Math.pow(2, exponent)) * 0x00800000) | 0;

            exponent += 127;
            if (exponent >= 0xFF) {
                exponent = 0xFF;
                significand = 0;
            } else if (exponent < 0) exponent = 0;

            bytes = bytes | (exponent << 23);
            bytes = bytes | (significand & ~(-1 << 23));
        break;
    }
    return bytes;
};

双精度(给定一个十进制值输出两个 32 位整数，二进制表示为大端序):

Double precision (given a decimal value outputs two 32-bit integers with the binary representation in big-endian order):

function toFloat64(value) {
    if ((byteOffset + 8) > this.byteLength) 
        throw "Invalid byteOffset: Cannot write beyond view boundaries.";

    var hiWord = 0, loWord = 0;
    switch (value) {
        case Number.POSITIVE_INFINITY: hiWord = 0x7FF00000; break;
        case Number.NEGATIVE_INFINITY: hiWord = 0xFFF00000; break;
        case +0.0: hiWord = 0x40000000; break;
        case -0.0: hiWord = 0xC0000000; break;
        default:
            if (Number.isNaN(value)) { hiWord = 0x7FF80000; break; }

            if (value <= -0.0) {
                hiWord = 0x80000000;
                value = -value;
            }

            var exponent = Math.floor(Math.log(value) / Math.log(2));
            var significand = Math.floor((value / Math.pow(2, exponent)) * Math.pow(2, 52));

            loWord = significand & 0xFFFFFFFF;
            significand /= Math.pow(2, 32);

            exponent += 1023;
            if (exponent >= 0x7FF) {
                exponent = 0x7FF;
                significand = 0;
            } else if (exponent < 0) exponent = 0;

            hiWord = hiWord | (exponent << 20);
            hiWord = hiWord | (significand & ~(-1 << 20));
        break;
    }

    return [hiWord, loWord];
};

对复制/粘贴中的任何错误表示歉意，代码也省略了对字节序的任何处理，尽管添加起来相当容易.

Apologies for any mistakes in copy/pasting, also the code ommits any handling of endianness, though it's fairly easy to add.

感谢大家提出的建议，但我最终还是靠自己解决了问题，因为我想尽可能避免循环以提高速度；它仍然不是非常快，但它会做=)

Thanks to everyone posting suggestions, but I ended up figuring out mostly on my own, as I wanted to avoid looping as much as possible for speed; it's still not exactly blazingly fast but it'll do =)

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