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?
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:
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), andFinally, 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.
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);
}
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