How to convert string to regex literal

What's the best way to escape an arbitrary std::wstring for use inside a regular expression? For example, convert you owe me $ to you owe me \$?

My scenario: I want to use a std::tr1::wregex to search for a whole word. So I want to do something like:

std::wstring RegexEscape(const std::wstring& inp)
{
    return ?????
}

bool ContainsWholeWord(const std::wstring& phrase, const std::wstring& word)
{
    std::tr1::wregex re开发者_如何转开发gex(std::wstring(L"\\b") + RegexEscape(word) + L"\\b");
    return std::tr1::regex_match(phrase, regex);
}

---

I don't know that it's the cleverest or the most efficient, but I use something like the following:

namespace {
bool
isMeta( char ch )
{
    static bool const meta[UCHAR_MAX] =
    {
        // ...
    };
    return meta[static_cast<unsigned char>( ch )];
}

std::string
sanitizeForRegEx( std::string const& original )
{
    std::string result;
    for ( std::string::const_iterator iter = original.begin();
            iter != original.end();
            ++ iter ) {
        if ( isMeta( *iter ) ) {
            result += '\\';
        result += *iter;
    }
    return result;
}

For wchar_t, I'd modify isMeta to return something like:

return ch >= 0 && ch < 128 && meta[ ch ];

The initialization of meta is a bit of a bore, and the exact values depend on the regular expressions used (or even the options if boost::regex is used).

---

Well, that's quite simple! Just use a regular expression to do that!

std::wstring szTmp; // some string with $, (, ...
std::wregex rgx_Meta( LR"(([\^\$\\\.\*\+\?\(\)\[\]\{\}\|]))" );
std::wstring strEscaped( std::regex_replace( szTmp, rgx_Meta, LR"(\$1)" ) );

This will replace all special character like '$' with '\$'.