打印Python和C++双倍时的精度差异

Question

我现在很惊讶：

<强> C++ 11

#include <iostream>
#include <iomanip>
#include <limits>

int main()
{
  double d = 1.305195828773568;
  std::cout << std::setprecision(std::numeric_limits<double>::max_digits10) << d << std::endl;
  // Prints  1.3051958287735681
}

Python

^{pr2}$ 什么事，为什么C++中有1个额外的值？在

到目前为止，我认为C++和Python在引擎盖下使用相同的64位IEEE双倍；两种格式函数都应该打印完整的精度。在

Answer 1

摘自对this question的回答：

IEEE 754 floating point is done in binary. There's no exact conversion from a given number of bits to a given number of decimal digits. 3 bits can hold values from 0 to 7, and 4 bits can hold values from 0 to 15. A value from 0 to 9 takes roughly 3.5 bits, but that's not exact either.

An IEEE 754 double precision number occupies 64 bits. Of this, 52 bits are dedicated to the significand (the rest is a sign bit and exponent). Since the significand is (usually) normalized, there's an implied 53rd bit.

Now, given 53 bits and roughly 3.5 bits per digit, simple division gives us 15.1429 digits of precision. But remember, that 3.5 bits per decimal digit is only an approximation, not a perfectly accurate answer.

您提供的15位数字后面的这个奇怪的.1429可能是添加的1的罪魁祸首。在

值得一提的是，Python在他们的网站上写了以下内容：

Historically, the Python prompt and built-in repr() function would choose the one with 17 significant digits, 0.10000000000000001. Starting with Python 3.1, Python (on most systems) is now able to choose the shortest of these and simply display 0.1.