哪个是IEEE 754浮点数不能精确表示的第一个整数？ [英] Which is the first integer that an IEEE 754 float is incapable of representing exactly?

问题描述

$ b

  float f0 = 0为了清楚起见，如果我使用的是实现IEE 754浮点数的语言， 。F; 
 float f1 = 1.f; 
  

...然后打印出来，我会得到0.0000和1.0000。 / p>

但是IEEE 754并不能代表真实线路上的所有数字。接近于零，差距很小;因此，我的问题是：对于IEEE 754浮点数，这是第一个（最接近于零）的浮点数，整数不能被精确表示？我现在只关心32位浮点数，尽管如果有人给它的话我会有兴趣听到64位的答案！

我认为这将如计算2 of_mantissa 并加1一样简单，其中 of_mantissa 是标准暴露的位数。我为我的机器上的32位浮点数（MSVC ++，Win64）做了这个工作，看起来不错。尾数比特+ 1 + 1

指数中的+1（尾数比特+ 1）是因为如果尾数包含 abcdef ... 它代表的数字实际上是 1.abcdef ...×2 ^ e ，提供了一个额外的隐含位精度。

对于 float ，它是16,777,217（2 24 + / + 1）。

对于 double ，它是9,007,199,254,740,993（2 ⁵³ + 1）。

 >>> 9007199254740993.0 
 9007199254740992 
  

For clarity, if I'm using a language that implements IEE 754 floats and I declare:

float f0 = 0.f;
float f1 = 1.f;

...and then print them back out, I'll get 0.0000 and 1.0000 - exactly.

But IEEE 754 isn't capable of representing all the numbers along the real line. Close to zero, the 'gaps' are small; as you get further away, the gaps get larger.

So, my question is: for an IEEE 754 float, which is the first (closest to zero) integer which cannot be exactly represented? I'm only really concerned with 32-bit floats for now, although I'll be interested to hear the answer for 64-bit if someone gives it!

I thought this would be as simple as calculating 2^{bits_of_mantissa} and adding 1, where bits_of_mantissa is how many bits the standard exposes. I did this for 32-bit floats on my machine (MSVC++, Win64), and it seemed fine, though.

解决方案

2^{mantissa bits + 1} + 1

The +1 in the exponent (mantissa bits + 1) is because, if the mantissa contains abcdef... the number it represents is actually 1.abcdef... × 2^e, providing an extra implicit bit of precision.

For float, it is 16,777,217 (2²⁴ + 1).
For double, it is 9,007,199,254,740,993 (2⁵³ + 1).

>>> 9007199254740993.0
9007199254740992

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