avoiding the first newline in a C++11 raw string literal?

Question

You can get a pointer to the 2nd character – skipping the leading newline – by adding 1 to the const char* to which the string literal is automatically converted:

    some_code();
    std::string text = 1 + R"(
This is the first line.
This is the second line.
This is the third line.
)";
    more_code();

IMHO, the above is flawed in breaking with the indentation of the surrounding code. Some languages provide a built-in or library function that does something like:

removes an empty leading line, and
looks at the indentation of the second line and removes the same amount of indentation from all further lines

That allows usage like:

some_code();
std::string text = unindent(R"(
    This is the first line.
    This is the second line.
    This is the third line.
    )");
more_code();

Writing such a function is relatively simple…

std::string unindent(const char* p)
{
    std::string result;
    if (*p == '\n') ++p;
    const char* p_leading = p;
    while (std::isspace(*p) && *p != '\n')
        ++p;
    size_t leading_len = p - p_leading;
    while (*p)
    {
        result += *p;
        if (*p++ == '\n')
        {
            for (size_t i = 0; i < leading_len; ++i)
                if (p[i] != p_leading[i])
                    goto dont_skip_leading;
            p += leading_len;
        }
      dont_skip_leading: ;
    }
    return result;
}

(The slightly weird p_leading[i] approach is intended to make life for people who use tabs and spaces no harder than they make it for themselves ;-P, as long as the lines start with the same sequence.)

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