好的,我已经想出解决方案了,接下来分享一下我针对单精度和双精度的解决方法。虽然我无法保证它们100%符合标准,但是它们不需要循环,似乎可以正常工作:
单精度(给定十进制值,输出32位大端单精度整数二进制表示):
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,并按大端序列的二进制表示):
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];
};
抱歉如果复制/粘贴中有任何错误,代码也省略了对字节序的处理,但很容易添加。
感谢所有提出建议的人,但我最终大部分是自己想出来的,因为我想尽可能避免循环以提高速度;虽然它仍不是非常快,但已经足够了=)
toString(2)
有帮助吗?(42).toString(2)
- MohsenNumber.toExponential
有一些用处。 - Paul S.