strange gcc warning

strange gcc warning

[Shanks]

Hi,

Sample program: b.c
#include <stdio.h>
#include <errno.h>

void
my_print(int errno)
{
    printf("ERRNO: %d\n", errno);
}

int
main()
{
    int a = 10;
    errno = 0;
    my_print(a);
}

On compiling this code i get the warning:
b.c: In function `main':
b.c:15: warning: passing arg 1 of `my_print' makes pointer from integer without
a cast

On running the code i get a seg. fault. So i did some gdb'ing and got
(gdb) run 
Starting program: ./b
Breakpoint 1, my_print (__errno_location=0x19) <<<<<<<<
    at b.c, line...
(gdb)

From above i see that instead of getting a plain int its go the address.
But i would like it if someone explained it to me better.

Thanks in advance.
-S

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Re: strange gcc warning

[Shanks]

--- Elias Athanasopoulos <eathan@otenet.gr> wrote:
> On Mon, Jan 13, 2003 at 09:57:55AM -0800, Shanks wrote:
> > my_print(int errno)
> 
> errno is defined in errno.h. You shouldn't use a new variable with this
> name.
Yes i am aware of errno being defined in errno.h
But isn't my definition local to my func.
Any technical reason why i am getting such a gdb output(pasted in prev mail)

> Elias
-Shanks

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Re: strange gcc warning

[Steven Smith]

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> void
> my_print(int errno)
Defining a parameter to shadow a global variable is always a bad idea,
but it's an especially bad idea for errno. In multi-threaded programs,
and programs compiled with a certain set of flags, errno is actually a
preprocessor macro.  Usually, when using glibc, errno expands to
(*__errno_location()), and so your prototype expands to void
my_print(int (*__errno_location())).  gcc interprets this as a pointer
to a function returning an int pointer, which was probably not what
you expected.

> {
>     printf("ERRNO: %d\n", errno);
Now, this expands to ``printf("ERRNO: %d\n", (*__errno_location()));".
__errno_location is a function pointer returning an int *, and so
this calls the first argument, and then dereferences the returned
value.

> }
> 
> int
> main()
> {
>     int a = 10;
>     errno = 0;
>     my_print(a);
a is now implicitly converted to a pointer to function returning int
*.  When the code is run, my_print tries to call a function at address
10 in memory, and crashes.

> }

Steven Smith,
sos22@cam.ac.uk.

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Re: strange gcc warning

[Elias Athanasopoulos]

On Mon, Jan 13, 2003 at 09:57:55AM -0800, Shanks wrote:
> my_print(int errno)

errno is defined in errno.h. You shouldn't use a new variable with this
name.

Elias

-- 
http://gnewtellium.sourceforge.net			MP3 is not a crime.	

Re: strange gcc warning

[Marius Nita]

On Mon, Jan 13, 2003 at 10:07:50AM -0800, Shanks wrote:
> > On Mon, Jan 13, 2003 at 09:57:55AM -0800, Shanks wrote:
> > > my_print(int errno)
> > 
> > errno is defined in errno.h. You shouldn't use a new variable with this
> > name.
> Yes i am aware of errno being defined in errno.h
> But isn't my definition local to my func.
> Any technical reason why i am getting such a gdb output(pasted in prev mail)

Steven Smith already explained everything in detail...

The reason 'locality' doesn't matter here is that errno is a macro. It
actually gets substituted by the preprocessor before the compiler ever
gets the chance to parse your program. It's like saying:

#define foo *bar

int print_foo(int foo) {
  printf("%d\n", foo);
}

your function will become

int print_foo(int *bar) {
  printf("%d\n", *bar);
}

so saying print_foo(9) is obviously a segfault error: you'll end up
dereferencing address 9, which is most likely not what you want :)

Re: strange gcc warning

[Shanks]

Thanks Steven for the clean and detailed explanation.
It all makes sense now... 

-Shanks.

--- Steven Smith <sos22@cam.ac.uk> wrote:
> > void
> > my_print(int errno)
> Defining a parameter to shadow a global variable is always a bad idea,
> but it's an especially bad idea for errno. In multi-threaded programs,
> and programs compiled with a certain set of flags, errno is actually a
> preprocessor macro.  Usually, when using glibc, errno expands to
> (*__errno_location()), and so your prototype expands to void
> my_print(int (*__errno_location())).  gcc interprets this as a pointer
> to a function returning an int pointer, which was probably not what
> you expected.
> 
> > {
> >     printf("ERRNO: %d\n", errno);
> Now, this expands to ``printf("ERRNO: %d\n", (*__errno_location()));".
> __errno_location is a function pointer returning an int *, and so
> this calls the first argument, and then dereferences the returned
> value.
> 
> > }
> > 
> > int
> > main()
> > {
> >     int a = 10;
> >     errno = 0;
> >     my_print(a);
> a is now implicitly converted to a pointer to function returning int
> *.  When the code is run, my_print tries to call a function at address
> 10 in memory, and crashes.
> 
> > }
> 
> Steven Smith,
> sos22@cam.ac.uk.

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Re: strange gcc warning

[Govind Raghuram]

Your definition will remain local as long as you dont include errno.h



                                              Placed At :


Shanks <mshanks79@yahoo.co.in> on 01/13/2003 11:37:50 PM

To:   Elias Athanasopoulos <eathan@otenet.gr>
cc:   linux-c-programming@vger.kernel.org (bcc: Govind Raghuram/Satyam)

Subject:  Re: strange gcc warning




--- Elias Athanasopoulos <eathan@otenet.gr> wrote:
> On Mon, Jan 13, 2003 at 09:57:55AM -0800, Shanks wrote:
> > my_print(int errno)
>
> errno is defined in errno.h. You shouldn't use a new variable with this
> name.
Yes i am aware of errno being defined in errno.h
But isn't my definition local to my func.
Any technical reason why i am getting such a gdb output(pasted in prev mail)

> Elias
-Shanks

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