Convert wchar_t to int

Question

how can I convert a wchar_t ('9') to a digit in the form of an int (9)?

I have the following code where I check whether or not peek is a digit:

if (iswdigit(peek)) {
    // store peek as numeric
}

Can I just subtract '0' or is there some Unicode specifics I should worry about?

Answer 1

Look into the atoi class of functions: http://msdn.microsoft.com/en-us/library/hc25t012(v=vs.71).aspx

Especially _wtoi(const wchar_t *string); seems to be what you're looking for. You would have to make sure your wchar_t is properly null terminated, though, so try something like this:

if (iswdigit(peek)) {
    // store peek as numeric
    wchar_t s[2];
    s[0] = peek;
    s[1] = 0;
    int numeric_peek = _wtoi(s);
}

Answer 2

If the question concerns just '9' (or one of the Roman digits), just subtracting '0' is the correct solution. If you're concerned with anything for which iswdigit returns non-zero, however, the issue may be far more complex. The standard says that iswdigit returns a non-zero value if its argument is "a decimal digit wide-character code [in the current local]". Which is vague, and leaves it up to the locale to define exactly what is meant. In the "C" locale or the "Posix" locale, the "Posix" standard, at least, guarantees that only the Roman digits zero through nine are considered decimal digits (if I understand it correctly), so if you're in the "C" or "Posix" locale, just subtracting '0' should work.

Presumably, in a Unicode locale, this would be any character which has the general category Nd. There are a number of these. The safest solution would be simply to create something like (variables here with static lifetime):

wchar_t const* const digitTables[] =
{
    L"0123456789",
    L"\u0660\u0661\u0662\u0663\u0664\u0665\u0666\u0667\u0668\u0669",
    // ...
};

//!     \return
//!         wch as a numeric digit, or -1 if it is not a digit
int asNumeric( wchar_t wch )
{
    int result = -1;
    for ( wchar_t const* const* p = std::begin( digitTables );
            p != std::end( digitTables ) && result == -1;
            ++ p ) {
        wchar_t const* q = std::find( *p, *p + 10, wch );
        if ( q != *p + 10 ) {
            result = q - *p;
    }
    return result;
}

If you go this way:

1. you'll definitely want to download the UnicodeData.txt file from the Unicode consortium ("Uncode Character Database"—this page has a links to both the Unicode data file and an explination of the encodings used in it), and
2. possibly write a simple parser of this file to extract the information automatically (e.g. when there is a new version of Unicode)—the file is designed for simple programmatic parsing.

Finally, note that solutions based on ostringstream and istringstream (this includes boost::lexical_cast) will not work, since the conversions used in streams are defined to only use the Roman digits. (On the other hand, it might be reasonable to restrict your code to just the Roman digits. In which case, the test becomes if ( wch >= L'0' && wch <= L'9' ), and the conversion is done by simply subtracting L'0'— always supposing the the native encoding of wide character constants in your compiler is Unicode (the case, I'm pretty sure, of both VC++ and g++). Or just ensure that the locale is "C" (or "Posix", on a Unix machine).

EDIT: I forgot to mention: if you're doing any serious Unicode programming, you should look into ICU. Handling Unicode correctly is extremely non-trivial, and they've a lot of functionality already implemented.

Answer 3

You could use boost::lexical_cast:

const wchar_t c = '9';
int n = boost::lexical_cast<int>( c );

Answer 4

Despite MSDN documentation, a simple test suggest that not only ranger L'0'-L'9' returns true.

for(wchar_t i = 0; i < 0xFFFF; ++i)
{
    if (iswdigit(i))
    {
        wprintf(L"%d : %c\n", i, i);
    }
}

That means that L'0' subtraction probably won't work as you may expected.

Answer 5

For most purposes you can just subtract the code for '0'.

However, the Wikipedia article on Unicode numerials mentions that the decimal digits are represented in 23 separate blocks (including twice in Arabic).

If you are not worried about that, then just subtract the code for '0'.