Question

assuming i have a string containing hexadecimal digits where every 2 hex digits represents a character in the ASCII set and i need to convert the string containing hex digits back to its character equivalent

i found what i was looking for in this code:-

#include <algorithm>
#include <stdexcept>

std::string hex_to_string(const std::string& input)
{
  static const char* const lut = "0123456789ABCDEF";
  size_t len = input.length();
  if (len & 1) throw std::invalid_argument("odd length");

  std::string output;
  output.reserve(len / 2);
  for (size_t i = 0; i < len; i += 2)
  {
    char a = input[i];
    const char* p = std::lower_bound(lut, lut + 16, a);
    if (*p != a) throw std::invalid_argument("not a hex digit");

    char b = input[i + 1];
    const char* q = std::lower_bound(lut, lut + 16, b);
    if (*q != b) throw std::invalid_argument("not a hex digit");

    output.push_back(((p - lut) << 4) | (q - lut));
  }
  return output;
}

i am rather new to C++ and i could understand till the part output.push_back(((p – lut) << 4) | (q – lut));
suppose the string contains a hex value of 72 (which represents the char ‘r’ in ACSII) and just before the push_back operation for the output string, the value of p and lut would be:-
p = “789ABCDEF”
lut = “0123456789ABCDEF”

but, (p – lut) in this function is yielding 7 as a result. i don’t quite understand how this happens.??

Answer 1

That’s pointer arithmetic.

The value of p is not "7890ABCDEF".  Rather that’s the content stored at the address held in p.  Since p is a pointer, it’s value is an address.

lut points to element 0, p points to element 7 in the same array. Therefore p - lut is 7.

For any n, p + n is the same as &p[n], that is, the address of the n-th element. Here that fact is used in reverse.