Hi, if I have a range of say 000080-0007FF and I want to see if a char containing hex is within that range, how can I do this? Thanks a lot.
Example
char t = 0xd790;
if (t is within range of 000080-0007FF) // true
Thanks
views:
330answers:
4
+7
A:
wchar_t t = 0xd790;
if (t >= 0x80 && t <= 0x7ff) ...
In C++, characters are interchangeable with integers and you can compare their values directly.
Note that I used wchar_t, because the char data type can only hold values up to 0xFF.
Greg Hewgill
2008-11-02 20:50:11
+1
A:
@Greg, I tried it and I get warning: comparison is always false due to limited range of data type
Ok just read your fix
BobS
2008-11-02 20:53:17
It sounds like your compiler is set to interpret 'char' as a signed data type (which ranges from -128 to 127). Use wchar_t as I edited my answer.
Greg Hewgill
2008-11-02 20:54:29
Is there another way other than wchar_t?
BobS
2008-11-02 20:55:33
yes, use a short which is 16 bits. Actually you really should be using unsigned values. So unsigned short.
Brian R. Bondy
2008-11-02 21:04:26
+3
A:
unsigned short t = 0xd790;
if (t >= 0x80 && t <= 0x7ff) ...
Since a char has a max value of 0xFF you cannot use it to compare anything with more hex digits than 2.
Brian R. Bondy
2008-11-02 21:05:50
A:
Since hex on a computer is nothing more than a way to print a number (like decimal), you can also do your comparison with plain old base 10 integers.
if( (t >= 128) && (t <= 2047) ) { }
More readable.
dicroce
2008-11-02 21:16:51
No. I would disagree. The Hex values are more readable. You can see the bit patterns from the hex values.
Martin York
2008-11-02 21:50:34
Readability is entirely dependent on context. If the bit patterns are relevant to the problem at hand hex is more readable (presumably this is the OP's case). If the numbers represent something like normal counts of items, decimal is more readable.
Steve Fallows
2008-11-03 04:21:52