是否可以在 C++ 中用十六进制浮点值初始化浮点变量？

像这样:

double d = 0x011.1; // wrong!

最佳答案

技术规范P0245 Hexadecimal floating literals for C++已于 2016 年 2 月在佛罗里达州 jackson 维尔的 ISO C++ 标准委员会投票加入 C++17。

C99语言也有这个特性，兼容C++特性。

但是，正如 Lưu Vĩnh Phúc 所指出的那样的评论，语法 0x011.1 不是标准的一部分。二进制指数对于十六进制浮点文字是必需的。原因之一是为了避免 0x011.1F 中尾随 F 的歧义。是小数部分的十六进制数字 F 还是表示 float 的 float 后缀？
因此，在 p 后面附加一个正或负的十进制数，例如:0x011.1p0。

查看更具可读性的页面 floating literal page on cppreference.com .

0x | 0X hex-digit-sequence
0x | 0X hex-digit-sequence .
0x | 0X hex-digit-sequence(optional) . hex-digit-sequence

Hexadecimal digit-sequence representing a whole number without a radix separator. The exponent is never optional for hexadecimal floating-point literals: 0x1ffp10, 0X0p-1, 0x1.p0, 0xf.p-1, 0x0.123p-1, 0xa.bp10l

The exponent syntax for hexadecimal floating-point literal has the form
p | P exponent-sign(optional) digit-sequence

exponent-sign, if present, is either + or -

suffix, if present, is one of f, F, l, or L. The suffix determines the type of the floating-point literal:

• (no suffix) defines double
• f F defines float
• l L defines long double

另请参阅 GitHub 上的当前工作草案 C++17，§ 2.13.4 float 文字 章:https://github.com/cplusplus/draft/raw/master/papers/n4604.pdf

floating-literal:
  decimal-floating-literal
  hexadecimal-floating-literal
decimal-floating-literal:
  fractional-constant exponent-part_opt floating-suffix_opt
  digit-sequence exponent-part floating-suffix_opt
hexadecimal-floating-literal:
  hexadecimal-prefix hexadecimal-fractional-constant binary-exponent-part floating-suffix_opt
  hexadecimal-prefix hexadecimal-digit-sequence binary-exponent-part floating-suffix_opt
fractional-constant:
  digit-sequence_opt . digit-sequence
  digit-sequence .
hexadecimal-fractional-constant:
  hexadecimal-digit-sequence_opt . hexadecimal-digit-sequence
  hexadecimal-digit-sequence .
exponent-part:
  e sign_opt digit-sequence
  E sign_opt digit-sequence
binary-exponent-part:
  p sign_opt digit-sequence
  P sign_opt digit-sequence
sign: one of
  + -
digit-sequence:
  digit
  digit-sequence ’_opt digit
floating-suffix: one of
  f l F L

¹ A floating literal consists of an optional prefix specifying a base, an integer part, a radix point, a fraction part, an e, E, p or P, an optionally signed integer exponent, and an optional type suffix. The integer and fraction parts both consist of a sequence of decimal (base ten) digits if there is no prefix, or hexadecimal (base sixteen) digits if the prefix is 0x or 0X. The literal is a decimal floating literal in the former case and a hexadecimal floating literal in the latter case. Optional separating single quotes in a digit-sequence or hexadecimal- digit-sequence are ignored when determining its value. [ Example: The literals 1.602’176’565e-19 and 1.602176565e-19 have the same value. — end example ] Either the integer part or the fraction part (not both) can be omitted. Either the radix point or the letter e or E and the exponent (not both) can be omitted from a decimal floating literal. The radix point (but not the exponent) can be omitted from a hexadecimal floating literal. The integer part, the optional radix point, and the optional fraction part, form the significand of the floating literal. In a decimal floating literal, the exponent, if present, indicates the power of 10 by which the significand is to be scaled. In a hexadecimal floating literal, the exponent indicates the power of 2 by which the significand is to be scaled. [ Example: The literals 49.625 and 0xC.68p+2 have the same value. — end example ] If the scaled value is in the range of representable values for its type, the result is the scaled value if representable, else the larger or smaller representable value nearest the scaled value, chosen in an implementation-defined manner. The type of a floating literal is double unless explicitly specified by a suffix. The suffixes f and F specify float, the suffixes l and L specify long double. If the scaled value is not in the range of representable values for its type, the program is ill-formed.

作为unwind建议，您可以使用strtof() .以下代码段解码十六进制浮点文字(无 C++17):

#include <iostream>
#include <cstdlib>
#include <cstdio>

int main(int argc, char *argv[])
{
  if (argc != 2)
  {
    std::cout <<"Usage: "<< argv[0] <<" 0xA.Bp-1  => Decode hexfloat" "\n";
    return 1;
  }

  long double l;
  double      d;
  float       f;

  std::cout <<"Decode floating point hexadecimal = "<< argv[1];
  //std::istringstream(argv[1]) >> std::hexfloat >> d;
  l = std::strtold(argv[1],NULL); if(errno == ERANGE) std::cout << "\n" "std::strtold() range error";
  d = std::strtod (argv[1],NULL); if(errno == ERANGE) std::cout << "\n" "std::strtod() range error";
  f = std::strtof (argv[1],NULL); if(errno == ERANGE) std::cout << "\n" "std::strtod() range error";

  std::cout <<"\n"  "long double = "<< std::defaultfloat << l <<'\t'<< std::hexfloat << l
            <<"\n"  "double      = "<< std::defaultfloat << d <<'\t'<< std::hexfloat << d
            <<"\n"  "float       = "<< std::defaultfloat << f <<'\t'<< std::hexfloat << f <<'\n';
}

关于c++ - 十六进制浮点字面量，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21526752/