[英]How to convert an IEEE single precision floating point to a decimal value
So I am trying to convert 46bfc000 (which is a floating-point number in IEEE single precision) into a decimal value.所以我试图将 46bfc000(IEEE 单精度中的浮点数)转换为十进制值。
I can get a approximate value, but not the exact value.我可以得到一个近似值,但不能得到准确值。 So here is my work for my approximate value:
所以这是我的近似值的工作:
1) Convert into binary: 0100 0110 1011 1111 1100 0000 0000 0000 1)转换成二进制: 0100 0110 1011 1111 1100 0000 0000 0000
2) Find b-exp : 141-127 2)找到 b-exp : 141-127
3) Convert what is after the decimal value : 2^-1 + 2^-5... = .552726746 3)转换十进制值后的内容:2^-1 + 2^-5... = .552726746
4) Now follow this equation format : (1)sign bit * (1. + value in step 3) * 2^b-exp 4)现在遵循这个等式格式:(1)符号位*(1。+步骤3中的值)* 2^b-exp
5) Calculate : +1 X (1.5527226746) X 2^14 = 25439.87501 5)计算:+1 X (1.5527226746) X 2^14 = 25439.87501
Now I know that the exact value is: 24544. But I am wondering if there is a way for me to get the exact number, or is it impossible to convert a IEEE single precision binary to a decimal value?现在我知道确切的值是:24544。但我想知道是否有办法让我得到确切的数字,或者是否不可能将 IEEE 单精度二进制转换为十进制值?
I have figured out the equation to get out the exact number of the binary representation, it is: sign * 2^b-exp * mantissa我已经找到了得到二进制表示的确切数字的等式,它是:符号 * 2^b-exp * 尾数
Edit: To get the right mantissa, you need to ONLY calculate it starting at the fractional part of the binary.编辑:要获得正确的尾数,您只需要从二进制的小数部分开始计算它。 So for example, if your fractional is 011 1111...
例如,如果您的小数是 011 1111 ...
Then you would do (1*2^-0) + (1*2^-1) + (1*2^-2)...然后你会做 (1*2^-0) + (1*2^-1) + (1*2^-2)...
Keep doing this for all the numbers and you'll get your mantissa.继续对所有数字执行此操作,您将获得尾数。
Instead of calculating all those bits behind the comma, which is heck of a job, IMO, just scale everything by 2^23 and subtract 23 more from the exponent for compensation.而不是计算逗号后面的所有那些位,这是一项非常艰巨的工作,IMO,只需将所有内容按 2^23 缩放并从指数中减去 23 以进行补偿。
This is explained in my article about floating point for Delphi .这在我关于 Delphi 的浮点数的文章中进行了解释。
First decode:第一次解码:
0 - 1000 1101 - 011 1111 1100 0000 0000 0000
Insert hidden bit:插入隐藏位:
0 - 1000 1101 - 1011 1111 1100 0000 0000 0000
In hex:十六进制:
0 - 8D - BFC000
0x8D = 141
, minus bias of 127
, that becomes 14
. 0x8D = 141
,减去127
偏差,变成14
。
I like to scale things, so the calculation is:我喜欢缩放东西,所以计算是:
sign * full_mantissa * (exp - bias - len)
where full_mantissa is the mantissa, including hidden bit, as integer ;其中 full_mantissa 是尾数,包括隐藏位,为整数; bias = 127 and len = 23 (the number of mantissa bits).
偏差 = 127 和 len = 23(尾数位数)。
So then it becomes:那么就变成了:
1 * 0xBFC000 * 2^(14-23) = 0xBFC000 / 0x200 = 0x5FE0 = 24544
because 2^(14-23) = 2^-9 = 1 / 2^9 = 1 / 0x200
.因为
2^(14-23) = 2^-9 = 1 / 2^9 = 1 / 0x200
。
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