如何无损地将双精度转换为字符串并返回八度 [英] How to lossless convert a double to string and back in Octave

问题描述

将双精度值保存到字符串时会损失一些精度.即使您使用非常多的数字，转换也可能不可逆，即如果您将双精度 x 转换为字符串 sx 然后再转换回来，您将得到一个数字 x' 可能不等于 x.这可能会导致一些问题，例如在检查一组测试中的差异时.一种可能性是使用二进制形式(例如本机二进制形式或 HDF5)，但我想将数字存储在文本文件中，因此我需要转换为字符串.我有一个可行的解决方案，但我问是否有一些标准或更好的解决方案.

When saving a double to a string there is some loss of precision. Even if you use a very large number of digits the conversion may not be reversible, i.e. if you convert a double x to a string sx and then convert back you will get a number x' which may not be bitwise equal to x. This may cause some problem for instance when checking for differences in a battery of tests. One possibility is to use binary form (for instance the native Binary form, or HDF5) but I want to store the number in a text file, so I need a conversion to a string. I have a working solution but I ask if there is some standard for this or a better solution.

在 C/C++ 中，您可以将 double 转换为诸如 char* 之类的整数类型，然后使用 printf("%02x",c[j]).然后例如 PI 将被转换为长度为 16 的字符串:54442d18400921fb.问题在于，如果您阅读六进制，您将不知道它是哪个数字.所以我会对一些组合感兴趣，例如 pi -> 3.14{54442d18400921fb}.第一部分是数字的(可能是低精度)十进制表示(通常我会使用 "%g" 输出转换)，大括号中的字符串是无损的十六进制表示.

In C/C++ you could cast the double to some integer type like char* and then convert each byte to an hexa of length 2 with printf("%02x",c[j]). Then for instance PI would be converted to a string of length 16: 54442d18400921fb. The problem with this is that if you read the hexa you don get any idea of which number it is. So I would be interested in some mix for instance pi -> 3.14{54442d18400921fb}. The first part is a (probably low precision) decimal representation of the number (typically I would use a "%g" output conversion) and the string in braces is the lossless hexadecimal representation.

我将代码作为答案传递

推荐答案

按照帖子中已经提出的想法，我写了以下功能，似乎有效.

Following the ideas already suggested in the post I wrote the following functions, that seem to work.

function s = dbl2str(d);
  z = typecast(d,"uint32");
  s = sprintf("%.3g{%08x%08x}\n",d,z);
endfunction

function d = str2dbl(s);

  k1 = index(s,"{");
  k2 = index(s,"}");
  ## Check that there is a balanced {} or not at all                                      
  assert((k1==0) == (k2==0));

  if k1>0; assert(k2>k1); endif
  if (k1==0);
     ## If there is not {hexa} part convert with loss
     d = str2double(s); 
  else  
    ## Convert lossless
    ss = substr(s,k1+1,k2-k1-1);
    z = uint32(sscanf(ss,"%8x",2));
    d = typecast(z,"double");
  endif 
endfunction

那我有

>> spi=dbl2str(pi)
spi = 3.14{54442d18400921fb}

>> pi2 = str2dbl(spi)
pi2 =  3.1416
>> pi2-pi
ans = 0
>> snan = dbl2str(NaN)
snan = NaN{000000007ff80000}

>> nan1 = str2dbl(snan)
nan1 =  NaN

进一步的改进是使用其他类型的编码，例如实例 Base64(如@CrisLuengo 在评论中所建议的那样)将二进制部分的长度从 16 字节减少到 11 字节.

A further improvement would be to use other type of enconding, for instance Base64 (as suggested by @CrisLuengo in a comment) that would reduce the length of the binary part from 16 to 11 bytes.

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