«

ll13.c:104: warning: implicit declaration of function 'isascii'
ll13.c:104: warning: nested extern declaration of 'isascii'

ll13.c:104 if (isascii(p[0]))

Oddly enough I was using isblank() before I changed it to isascii() in
that statement and there was no error!
I have "#include <ctype.h>" so I don't understand this error. How can
I fix this?

Zach

"Zach" <> wrote in message
news:918f2354-9adc-498d-9d01-...
> ll13.c:104: warning: implicit declaration of function 'isascii'
> ll13.c:104: warning: nested extern declaration of 'isascii'
>
> ll13.c:104 if (isascii(p[0]))
>
> Oddly enough I was using isblank() before I changed it to isascii() in
> that statement and there was no error!
> I have "#include <ctype.h>" so I don't understand this error. How can
> I fix this?

That's not a standard C function. Your compiler setting may be preventing
extensions to standard C headers like ctype.h from including non-standard
functions. Why are you using it?



Brian
--
Day 652 of the "no grouchy usenet posts" project.
Current music playing: None.

On Nov 18, 7:13*pm, "Default User" <defaultuse...@yahoo.com> wrote:
>
> That's not a standard C function. Your compiler setting may be preventing
> extensions to standard C headers like ctype.h from including non-standard
> functions. Why are you using it?

But gcc has no problem when I use other functions from ctype.h like
isblank(). I need to check if a character (looping through a string)
is valid or not. Before I was using "!isblank()" but isascii() seems
to recognize more characters. This is part of a function that counts
how many tokens are in a string.

Zach

Zach <> writes:
> On Nov 18, 7:13Â*pm, "Default User" <defaultuse...@yahoo.com> wrote:
>> That's not a standard C function. Your compiler setting may be preventing
>> extensions to standard C headers like ctype.h from including non-standard
>> functions. Why are you using it?
>
> But gcc has no problem when I use other functions from ctype.h like
> isblank(). I need to check if a character (looping through a string)
> is valid or not. Before I was using "!isblank()" but isascii() seems
> to recognize more characters. This is part of a function that counts
> how many tokens are in a string.

isblank() is standard. isascii() isn't.

You're probably telling gcc to operate in standard-conforming mode,
which means that isascii() isn't going to be visible. If you
really want to use it, you'll need to turn off whatever options
you're using that are causing gcc to disable it.

But again, why are you using isascii()? Since it returns true for
any 7-bit character that fits into the 7-bit ASCII character set,
*including the space character*, I doubt that it's going to be
useful for your purpose.

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On 19 nov, 01:01, Zach <net...@gmail.com> wrote:
> ll13.c:104: warning: implicit declaration of function 'isascii'
> ll13.c:104: warning: nested extern declaration of 'isascii'
>
> ll13.c:104 if (isascii(p[0]))
>
> Oddly enough I was using isblank() before I changed it to isascii() in
> that statement and there was no error!
> I have "#include <ctype.h>" so I don't understand this error. How can
> I fix this?
>
> Zach

Is it possible you might have thought of isalnum() or isalpha()
instead?
Hans

On Nov 18, 8:04*pm, Keith Thompson <ks...@mib.org> wrote:
>
> isblank() is standard. *isascii() isn't.
>
> You're probably telling gcc to operate in standard-conforming mode,
> which means that isascii() isn't going to be visible. *If you
> really want to use it, you'll need to turn off whatever options
> you're using that are causing gcc to disable it.
>
> But again, why are you using isascii()? *Since it returns true for
> any 7-bit character that fits into the 7-bit ASCII character set,
> *including the space character*, I doubt that it's going to be
> useful for your purpose.

Ah, thanks Ken. Will stick with isblank().

Zach

Zach <> writes:
> On Nov 18, 8:04Â*pm, Keith Thompson <ks...@mib.org> wrote:
>>
>> isblank() is standard. Â*isascii() isn't.
>>
>> You're probably telling gcc to operate in standard-conforming mode,
>> which means that isascii() isn't going to be visible. Â*If you
>> really want to use it, you'll need to turn off whatever options
>> you're using that are causing gcc to disable it.
>>
>> But again, why are you using isascii()? Â*Since it returns true for
>> any 7-bit character that fits into the 7-bit ASCII character set,
>> *including the space character*, I doubt that it's going to be
>> useful for your purpose.
>
> Ah, thanks Ken. Will stick with isblank().

You're welcome. Who's Ken?

--
Keith Thompson (The_Other_Keith) kst- <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

On Thu, 18 Nov 2010 17:04:36 -0800, Keith Thompson wrote:

> You're probably telling gcc to operate in standard-conforming mode,
> which means that isascii() isn't going to be visible. If you
> really want to use it, you'll need to turn off whatever options
> you're using that are causing gcc to disable it.

Normally, you'd just define the relevant feature-test macros. The
isascii(3) manpage says:

Feature Test Macro Requirements for glibc (see feature_test_macros(7)):

isascii():
_BSD_SOURCE || _SVID_SOURCE || _XOPEN_SOURCE

Or use _GNU_SOURCE, which enables all of these and then some. But
if you're using -ansi/-std=, you might be trying to avoid that.

On Nov 19, 11:36*am, Keith Thompson <ks...@mib.org> wrote:
>
>
> You're welcome. *Who's Ken?

Sorry, must have been thinking of Ken Thompson

Zach