Re: Weird warning about data type range

From:
Jerry Coffin <jerryvcoffin@yahoo.com>
Newsgroups:
comp.lang.c++
Date:
Sun, 30 Aug 2009 12:27:35 -0600
Message-ID:
<MPG.2504720c57d590c298979b@news.sunsite.dk>
In article <ad7ec9cc-6e85-4d12-8ebf-6072d20e2617
@z4g2000prh.googlegroups.com>, digital_puer@hotmail.com says...

Thanks for everyone's help.

What I am really trying to do is to test if
characters in a std::string are alphanumeric
characters in the Latin-1 encoding. I have
the following:

#define IS_ALPHANUMERIC(x) ( \
    ((x) >= 48 && (x) <= 57 ) || \
    ((x) >= 65 && (x) <= 90 ) || \
    ((x) >= 97 && (x) <= 122) || \
    ((x) >= 192 && (x) <= 214) || \
    ((x) >= 216 && (x) <= 246) || \
    ((x) >= 248 && (x) <= 255) )

string s = getLatin1Text();
int len = s.size();
for (int i = 0; i < len; i++)
{
  if (! IS_ALPHANUMERIC(s.at(i)))
  {
     ...
  }
}


If I had to do this, I think I'd use something like this:

#include <climits>
#include <vector>
#include <algorithm>

struct alphanumeric_table {
    std::vector<bool> table;
public:
#define elements(r) (sizeof(r)/sizeof(r[0]))

    alphanumeric_table() : table(UCHAR_MAX+2, false) {
        static const int ranges[] = {
            48, 57,
            65, 90,
            97, 122,
            192, 214,
            216, 246,
            248, 255
        };

        for (int i=0; i<elements(ranges); i+=2)
            std::fill(table.begin()+ranges[i]+1,
                      table.begin()+ranges[i+1]+2,
                      true);
    }

    bool operator[](int n) { return table[unsigned char(n+1)]; }
} alpha_table;

inline bool is_alphanumeric(int n) {
    return alpha_table[n];
}

This assumes that EOF is -1. Technically this isn't required (any
negative value is allowed) but it's extremely common -- to the point
that I'm not sure I've ever seen or heard of it actually having any
other value. In any case, the reason for the "+1"in most places is to
get a range from 0 through the maximum, so we can use it directly as
an index into the vector.

--
    Later,
    Jerry.

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