在Javascript中将浮点数转换为二进制表示形式的最简单方法是什么? (例如1.0 - > 0x3F800000)。
我试图手动完成,这在一定程度上(通常的数字)有效,但是对于非常大或非常小的数字(没有范围检查)和特殊情况(NaN,无限等)都会失败:
function floatToNumber(flt)
{
var sign = (flt < 0) ? 1 : 0;
flt = Math.abs(flt);
var exponent = Math.floor(Math.log(flt) / Math.LN2);
var mantissa = flt / Math.pow(2, exponent);
return (sign << 31) | ((exponent + 127) << 23) | ((mantissa * Math.pow(2, 23)) & 0x7FFFFF);
}
我是否重新发明轮子?
编辑:我改进了我的版本,现在它处理特殊情况。function assembleFloat(sign, exponent, mantissa)
{
return (sign << 31) | (exponent << 23) | (mantissa);
}
function floatToNumber(flt)
{
if (isNaN(flt)) // Special case: NaN
return assembleFloat(0, 0xFF, 0x1337); // Mantissa is nonzero for NaN
var sign = (flt < 0) ? 1 : 0;
flt = Math.abs(flt);
if (flt == 0.0) // Special case: +-0
return assembleFloat(sign, 0, 0);
var exponent = Math.floor(Math.log(flt) / Math.LN2);
if (exponent > 127 || exponent < -126) // Special case: +-Infinity (and huge numbers)
return assembleFloat(sign, 0xFF, 0); // Mantissa is zero for +-Infinity
var mantissa = flt / Math.pow(2, exponent);
return assembleFloat(sign, exponent + 127, (mantissa * Math.pow(2, 23)) & 0x7FFFFF);
}
我仍然不确定这是否100%正确,但似乎工作得很好。 (我仍然在寻找现有的实现)。
答案 0 :(得分:7)
新技术使这一过程变得简单,也可能更加向前兼容。我喜欢扩展内置原型,而不是每个人都这样做。所以随意修改以下代码到经典的程序方法:
(function() {
function NumberToArrayBuffer() {
// Create 1 entry long Float64 array
return [new Float64Array([this]).buffer];
}
function NumberFromArrayBuffer(buffer) {
// Off course, the buffer must be at least 8 bytes long, otherwise this is a parse error
return new Float64Array(buffer, 0, 1)[0];
}
if(Number.prototype.toArrayBuffer) {
console.warn("Overriding existing Number.prototype.toArrayBuffer - this can mean framework conflict, new WEB API conflict or double inclusion.");
}
Number.prototype.toArrayBuffer = NumberToArrayBuffer;
Number.prototype.fromArrayBuffer = NumberFromArrayBuffer;
// Hide this methods from for-in loops
Object.defineProperty(Number.prototype, "toArrayBuffer", {enumerable: false});
Object.defineProperty(Number.prototype, "fromArrayBuffer", {enumerable: false});
})();
(function() {
function NumberToArrayBuffer() {
// Create 1 entry long Float64 array
return new Float64Array([this.valueOf()]).buffer;
}
function NumberFromArrayBuffer(buffer) {
// Off course, the buffer must be ar least 8 bytes long, otherwise this is a parse error
return new Float64Array(buffer, 0, 1)[0];
}
if(Number.prototype.toArrayBuffer) {
console.warn("Overriding existing Number.prototype.toArrayBuffer - this can mean framework conflict, new WEB API conflict or double inclusion.");
}
Number.prototype.toArrayBuffer = NumberToArrayBuffer;
Number.fromArrayBuffer = NumberFromArrayBuffer;
// Hide this methods from for-in loops
Object.defineProperty(Number.prototype, "toArrayBuffer", {enumerable: false});
Object.defineProperty(Number, "fromArrayBuffer", {enumerable: false});
})();
var test_numbers = [0.00000001, 666666666666, NaN, Infinity, -Infinity,0,-0];
console.log("Conversion symethry test: ");
test_numbers.forEach(
function(num) {
console.log(" ", Number.fromArrayBuffer((num).toArrayBuffer()));
}
);
console.log("Individual bytes of a Number: ",new Uint8Array((666).toArrayBuffer(),0,8));
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答案 1 :(得分:5)
这是一个适用于我测试过的所有内容的函数,除了它不区分-0.0和+0.0。
它基于http://jsfromhell.com/classes/binary-parser的代码,但它专门用于32位浮点数并返回整数而不是字符串。我还对其进行了修改,使其更快,(稍微)更具可读性。
// Based on code from Jonas Raoni Soares Silva
// http://jsfromhell.com/classes/binary-parser
function encodeFloat(number) {
var n = +number,
status = (n !== n) || n == -Infinity || n == +Infinity ? n : 0,
exp = 0,
len = 281, // 2 * 127 + 1 + 23 + 3,
bin = new Array(len),
signal = (n = status !== 0 ? 0 : n) < 0,
n = Math.abs(n),
intPart = Math.floor(n),
floatPart = n - intPart,
i, lastBit, rounded, j, exponent;
if (status !== 0) {
if (n !== n) {
return 0x7fc00000;
}
if (n === Infinity) {
return 0x7f800000;
}
if (n === -Infinity) {
return 0xff800000
}
}
i = len;
while (i) {
bin[--i] = 0;
}
i = 129;
while (intPart && i) {
bin[--i] = intPart % 2;
intPart = Math.floor(intPart / 2);
}
i = 128;
while (floatPart > 0 && i) {
(bin[++i] = ((floatPart *= 2) >= 1) - 0) && --floatPart;
}
i = -1;
while (++i < len && !bin[i]);
if (bin[(lastBit = 22 + (i = (exp = 128 - i) >= -126 && exp <= 127 ? i + 1 : 128 - (exp = -127))) + 1]) {
if (!(rounded = bin[lastBit])) {
j = lastBit + 2;
while (!rounded && j < len) {
rounded = bin[j++];
}
}
j = lastBit + 1;
while (rounded && --j >= 0) {
(bin[j] = !bin[j] - 0) && (rounded = 0);
}
}
i = i - 2 < 0 ? -1 : i - 3;
while(++i < len && !bin[i]);
(exp = 128 - i) >= -126 && exp <= 127 ? ++i : exp < -126 && (i = 255, exp = -127);
(intPart || status !== 0) && (exp = 128, i = 129, status == -Infinity ? signal = 1 : (status !== status) && (bin[i] = 1));
n = Math.abs(exp + 127);
exponent = 0;
j = 0;
while (j < 8) {
exponent += (n % 2) << j;
n >>= 1;
j++;
}
var mantissa = 0;
n = i + 23;
for (; i < n; i++) {
mantissa = (mantissa << 1) + bin[i];
}
return ((signal ? 0x80000000 : 0) + (exponent << 23) + mantissa) | 0;
}