My ISDIGIT() Function has fallen and cant get up!!! 1

[BigTeeJay]

Greetings,
I appears that I "broke" my isdigit() function I was
using it to test arguments passed to a VERY simple (or
so I thought) command line DOS program. It was originally
evaluating input... but now appears to return 0 (which
according to my book & my debugging efforts means it
wasnt an integer) regardless of what value is passed to it.

Below is example code...

#include <iostream.h>
#include <stdlib.h>
#include <ctype.h>

int main(int argc, char *argv[])
{
	//process cmd line, see whats returned, currently always a 0 (zero)
	cout << endl;
	cout << &quot;ATOI (v): &quot; << atoi( argv[1] ) << endl;
	cout << &quot;ISDIG(v): &quot; << isdigit( atoi( argv[1] ) ) << endl << endl;

	//process a fixed string, see what happens (returns 0 for me)
	char *tstStr = &quot;3&quot;;
	cout << &quot;ATOI (s): &quot; << atoi( tstStr ) << endl;
	cout << &quot;ISDIG(s): &quot; << isdigit( atoi( tstStr ) ) << endl << endl;

	//pass a freaking INT, which isdigit still says isnt a digit!?!?!?
	int tstInt = 3;
	cout << &quot;VALUE(i): &quot; << tstInt << endl;
	cout << &quot;ISDIG(i): &quot; << isdigit( tstInt ) << endl << endl

	return 0;
}

...in desparation I even commented out the "endl"'s
being passed to cout, thinking there was some remote
chance that these were being evaluated by the isdigit
function too (I know... but, like I said... desparate).

This is driving me nuts!!!!!!! What am I missing here?
I am sure this is going to be something I am going to
want to crawl under a rock when someone points out an
blatent miss... but until then, I cant find it for the
life of me.

Regards,
Tj

 

[Guest_imported]

You misunderstand the isdigit(int). The argument here is an integer. If it is between '0' and '9', i.e, 48-57, it will return a non-zero integer, otherwise it will return 0.
try isdigit('3') to see what it returns.

 

[Zyrenthian]

isdigit is decieving. What it is really looking for is an ascii value so '0' '1' etc will work and in the case of

char *tstStr = "3";

if you call

isdigit(tstStr[0]) you should get a true value

Matt

 

[BigTeeJay]

argggggg..... I started thinking about that
last night, when I was looking at the code
and thought that digit doesnt necessarily
mean int... but was too tired to think straight

I thank you both for your assistance!

Regards,
Tj