numeric_limits<double>::digits10是什么意思

2022-01-07 00:00:00 c++ stl

numeric_limits::digits10 的确切含义是什么?stackoverflow 中的一些其他相关问题让我认为它是双精度的最大精度，但是

What is the precise meaning of numeric_limits::digits10? Some other related questions in stackoverflow made me think it is the maximum precision of a double, but

  • 当精度大于 17 ( == 2+numeric_limits::digits10) 时，以下原型开始工作(成功为真)
  • 使用STLPort，最后readDouble==infinity；使用微软的 STL，readDouble == 0.0.
  • 这个原型有什么意义吗:)?

这是原型:

#include <float.h>
#include <limits>
#include <math.h>
#include <iostream>
#include <iomanip>
#include <sstream>
#include <string>
int main(int argc, const char* argv[]) {
  std::ostringstream os;
  //int digit10=std::numeric_limits<double>::digits10; // ==15
  //int digit=std::numeric_limits<double>::digits; // ==53
  os << std::setprecision(17);
  os << DBL_MAX;
  std::cout << os.str();
  std::stringbuf sb(os.str());
  std::istream is(&sb);
  double readDouble=0.0;
  is >> readDouble;
  bool success = fabs(DBL_MAX-readDouble)<0.1;
}

推荐答案

numeric_limits::digits10 是可以不丢失的小数位数.

numeric_limits::digits10 is the number of decimal digits that can be held without loss.

例如 numeric_limits::digits10 是 2.这意味着无符号字符可以保持 0..99 而不会丢失.如果是 3，它可以容纳 0..999，但众所周知，它只能容纳 0..255.

For example numeric_limits<unsigned char>::digits10 is 2. This means that an unsigned char can hold 0..99 without loss. If it were 3 it could hold 0..999, but as we all know it can only hold 0..255.

本手册页有一个浮点数示例，其中(缩短时)表明

This manual page has an example for floating point numbers, which (when shortened) shows that

cout << numeric_limits<float>::digits10 <<endl;
float f = (float)99999999; // 8 digits
cout.precision ( 10 );
cout << "The float is; " << f << endl;

印刷品

6
The float is; 100000000

相关文章