IEEE 754 浮点数无法准确表示的第一个整数是哪一个？
为了清楚起见，如果我使用实现 IEE 754 浮点的语言并且我声明：
float f0 = 0.f;
float f1 = 1.f;
...然后将它们打印回来，我将准确地得到 0.0000 和 1.0000。
但 IEEE 754 无法表示实数线上的所有数字。接近于零，“差距”很小；当你离得更远时，差距就会变得更大。
所以，我的问题是：对于 IEEE 754 浮点数，这是第一个（最接近零）无法精确表示的整数？我现在只关心 32 位浮点数，尽管如果有人给出 64 位的答案，我将很感兴趣！
我认为这就像计算 2bits_of_mantissa 并加 1 一样简单，其中 bits_of_mantissa 是标准公开的位数。我在我的机器（MSVC++、Win64）上对 32 位浮点数执行了此操作，但看起来不错。
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 2bits_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.
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2尾数位 + 1 + 1
指数（尾数位+1）中的+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，提供额外的隐式精度。因此，第一个无法精确表示且将被舍入的整数是：
以下是 CPython 3.10 中的示例，它使用 64 位浮点数：
2mantissa bits + 1 + 1
The +1 in the exponent (mantissa bits + 1) is because, if the mantissa contains
abcdef...the number it represents is actually1.abcdef... × 2^e, providing an extra implicit bit of precision.Therefore, the first integer that cannot be accurately represented and will be rounded is:
Here's an example in CPython 3.10, which uses 64-bit floats:
n 位整数可表示的最大值为 2n-1。如上所述，
float的尾数精度为 24 位，这似乎意味着 224 不适合。但是。
指数范围内的2的幂可以精确地表示为1.0×2n，因此224可以拟合，因此
float的第一个不可表示的整数是 224+1。如上所述。再次。The largest value representable by an n bit integer is 2n-1. As noted above, a
floathas 24 bits of precision in the significand which would seem to imply that 224 wouldn't fit.However.
Powers of 2 within the range of the exponent are exactly representable as 1.0×2n, so 224 can fit and consequently the first unrepresentable integer for
floatis 224+1. As noted above. Again.