IsUnsigned() function?

[Cedric Ramsey]

In article <4700059 at m.cs.uiuc.edu> carroll at m.cs.uiuc.edu writes:
>
>/* Written 12:07 pm  Aug  2, 1990 by henry at zoo.toronto.edu in m.cs.uiuc.edu:comp.lang.c */
>In article <4700058 at m.cs.uiuc.edu> carroll at m.cs.uiuc.edu writes:
>>The problem was that "signed" was a valid type modifier, but "unsigned" wasn't.
>>If chars were (by default) unsigned, they could be made signed...
>
>Um, excuse me?
>/* End of text from m.cs.uiuc.edu:comp.lang.c */
>Woops. Change "signed" <-> "unsigned". Got to get some more sleep.


	Hello, I would like to throw in my two cents here, for what it's worth.
I though that chars were signed or unsigned for when used in expressions they
will be promoted to integer or unsigned as declared. Otherwise, who cares.
a char will be a char no matter what it's sign is. It will have the same
number of bits, which is the real problem; some chars are 8 bit others are 
7 some are 9, but thats another story. Anyway, that all that I have to say. 
As to the question, I don't know, I wasn't there. But if I didn't understand
the question I would have asked for clarification. Afterall, how can a 
programmer try to program a task without an understanding of what is 
to be done. That is from my perspective but then again I wasn't there and
don't know any of the underlying circumstances.