Escaped represention of std::string

Question

I'm executing the following code, have read this:

#include <iostream>
#include <boost/format.hpp>
#include <boost/algorithm/string.hpp>

struct CharacterEscaper
{
    template<typename FindResultT>
    std::string operator()(const FindResultT& Match) const
    {
        std::string s;
        for (typename FindResultT::const_iterator i = Match.begin();i != Match.end();i++) 
        {
            s += str(boost::format("\\x%02x") % static_cast<int>(*i));
        }
        return s;
    }
};

int main (int argc, char **argv)
{
    std::string s("start\x0aend");
    boost::find_format_all(s, boost::token_finder(!boost::is_print()), CharacterEscaper());
    std::cout << s << std::endl;
    return 0;
}

The output I'm getting is this:

start\xffffffaend

I was expecting the following:

start\x0aend

having read this:

http://en.cppreference.com/w/cpp/language/escape

What is correct?

score 1 · Answer 1 · answered Dec 14 '12 at 15:06

Two issues here:

The escaper converts the character value to int before formatting it. This means that, if char is signed, values above 0x7f will be sign-extended, giving the extra fs in the output. You can fix that by converting first to unsigned char, then perhaps to unsigned or int. (I don't know enough about the Boost formatter to know whether the second converion is necessary.)
A hexadecimal escape sequence will include as many hexadecimal digits as it can find, so in this case it will be \x0ae. One way to fix that is by breaking the string literal into two, which will be recombined during compilation: "start\x0a" "end"

score 0 · Answer 2 · edited Dec 14 '12 at 15:08

0

I think I have the answer. I should have written this: "start\x0a" "end"

edited Dec 14 '12 at 15:08

MSalters

173,980
10
155
350

answered Dec 14 '12 at 14:31

Baz

12,713
38
145
268

1

Well, that's one solution. You do get similar problems with char values above `\x7F` though (because those are really values below 0). With what you have now, the most simple solution is to replace the `` with ``. – Mr Lister Dec 14 '12 at 14:35
@Mr Lister How might I fix this? – Baz Dec 14 '12 at 14:37
1

That is because it should be a `static_cast`. – Bart van Ingen Schenau Dec 14 '12 at 15:04
@Bart Right. I knew I was on the right track, but I wasn't thinking straight. – Mr Lister Dec 14 '12 at 15:50

Escaped represention of std::string

2 Answers2