assignment makes pointer from integer without a cast?

assignment makes pointer from integer without a cast?

[Patrick]

Hello,

New to the list and to C. Read O'Reilly's Practical C Programming and
another C book and am now trying to evolve beyond helloworld.c. When I
compile the snippet below, gcc-3.4.2 on Fedora Core 3 spits out:
test2.c:11: warning: assignment makes pointer from integer without a
cast
test2.c:12: warning: passing arg 1 of `usage' from incompatible pointer
type

I just can't figure out why it does that and how to fix it since my
pointer/casting/C knowledge is definitely lacking. The snippet does work
ok. Anyone care to enlighten me?

#include <stdio.h>

void usage(int *prog_ptr)
{
	fprintf(stdout,"progname is: %s\n",*prog_ptr);
}
	
int main(int argc, char *argv[])
{
	int *prog_ptr;
	prog_ptr = basename (argv[0]);
	usage(&prog_ptr);
	return(0);
}

TIA,
Patrick

Re: assignment makes pointer from integer without a cast?

[Hossein Mobahi]

Patrick

Your program has a few incorrect assignments. Is
prog_ptr going to hold a string? I guess so because
you want to print program name. So "int *prog_ptr" is
wrong, use "char *prog_ptr" instead, i.e. an address
pointing to the first character of a given string. 

Moreover, you do not need to use ampersand to obtain
prog_ptr's address because it is already a pointer
itself not an ordinary variable.

Eventually, you must pass a pointer to fprintf when
you use %s. So the * is incorrect (*prog_ptr returns a
character only, the first character of your string,
which is not what you want).

I'd rewrite your program as follows:

#include <stdio.h>

void usage(char *prog_ptr) /* not int* */
{
	fprintf(stdout,"progname is: %s\n",prog_ptr); /* not
*prog_ptr */
}
	
int main(int argc, char *argv[])
{
	char *prog_ptr; /* not int * */
	prog_ptr = basename (argv[0]);
	usage(prog_ptr); /* not &prog_ptr */
	return(0);
}


--- Patrick <patrickC@puzzled.xs4all.nl> wrote:
> #include <stdio.h>
> 
> void usage(int *prog_ptr)
> {
> 	fprintf(stdout,"progname is: %s\n",*prog_ptr);
> }
> 	
> int main(int argc, char *argv[])
> {
> 	int *prog_ptr;
> 	prog_ptr = basename (argv[0]);
> 	usage(&prog_ptr);
> 	return(0);
> }



		
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Re: assignment makes pointer from integer without a cast?

[Francesco Gadaleta]

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> void usage(char *prog_ptr) /* not int* */
> {
> 	fprintf(stdout,"progname is: %s\n",prog_ptr); /* not
> *prog_ptr */
> }
> 	
> int main(int argc, char *argv[])
> {
> 	char *prog_ptr; /* not int * */
> 	prog_ptr = basename (argv[0]);
> 	usage(prog_ptr); /* not &prog_ptr */
> 	return(0);
> }
> 
> 


change this line:
prog_ptr = basename (argv[0]);

with this:
prog_ptr = (char*) basename (argv[0]);
-- 

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Re: assignment makes pointer from integer without a cast?

[Jan-Benedict Glaw]

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On Mon, 2004-12-13 17:13:54 +0100, Francesco Gadaleta <francesco@gadaleta.org>
wrote in message <20041213161354.GD1487@sweetdream>:
> > void usage(char *prog_ptr) /* not int* */
> > {
> > 	fprintf(stdout,"progname is: %s\n",prog_ptr); /* not
> > *prog_ptr */
> > }
> > 	
> > int main(int argc, char *argv[])
> > {
> > 	char *prog_ptr; /* not int * */
> > 	prog_ptr = basename (argv[0]);
> > 	usage(prog_ptr); /* not &prog_ptr */
> > 	return(0);
> > }
> change this line:
> prog_ptr = basename (argv[0]);
> 
> with this:
> prog_ptr = (char*) basename (argv[0]);

No, better include the proper header file <libgen.h> or <string.h>
(after #defining _GNU_SOURCE). This way, you don't force gcc into
guessing it's type. If you had compiled this with a C++ compiler (*), it
wouldn't even had accepted the code or wouldn't link it later on.

MfG, JBG
(*) Compiling not-too-complicated C code with a C++ compiler can reveal
some additional oddities that a pure C compiler wouldn't warn on. So if
you keep your code being compileable with both a C and a  C++ compiler
when you start a new project, this won't hurt (but probably pay back at
some time). (-> Right, I'm a C programmer and personally don't use C++
that often, because it's too hard to write "basic" stuff with it).

-- 
Jan-Benedict Glaw       jbglaw@lug-owl.de    . +49-172-7608481             _ O _
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how to link mqueue library

[Ron Michael Khu]

Hello.

I seem to have a problem linking the mqueue library....
What's the library name to be added to the "-l" option of cc/gcc?
If -lm is for the math library... -lcurses for ncurses,
what's the library name for mqueue? -lmqueue??

-Ron.

Re: how to link mqueue library

[Camillo Dell'mour]

Hi Ron
the math lib is named libm and the curses lib  libcurses

so the by the -l option in gcc you do not have to write "lib..."

I don't  who the mqueue libary is named, But I think libqueue so you shoult write -lqueue

ron wrots --
> 
> Hello.
> 
> I seem to have a problem linking the mqueue library....
> What's the library name to be added to the "-l" option of cc/gcc?
> If -lm is for the math library... -lcurses for ncurses,
> what's the library name for mqueue? -lmqueue??
> 
> -Ron.
> 


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Re: how to link mqueue library

[Ron Michael Khu]

Thanks camillo.

I already found the library;
it's librt.....


Camillo Dell'mour wrote:

>Hi Ron
>the math lib is named libm and the curses lib  libcurses
>
>so the by the -l option in gcc you do not have to write "lib..."
>
>I don't  who the mqueue libary is named, But I think libqueue so you shoult write -lqueue
>
>ron wrots --
>  
>
>>Hello.
>>
>>I seem to have a problem linking the mqueue library....
>>What's the library name to be added to the "-l" option of cc/gcc?
>>If -lm is for the math library... -lcurses for ncurses,
>>what's the library name for mqueue? -lmqueue??
>>
>>-Ron.
>>
>>    
>>
>
>
>__________________________________________________________
>Mit WEB.DE FreePhone mit hoechster Qualitaet ab 0 Ct./Min.
>weltweit telefonieren! http://freephone.web.de/?mc=021201
>
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