ioctl in Ubuntus 10.04 GCC

[BGraaf]

Hi,

is there anyone to help me using the ioctl-function. My code looks like this:

------


#include <stdio.h>
#include <fcntl.h>
#include <sys/ioctl.h>
#include <linux/dvb/dmx.h>
#include <linux/dvb/frontend.h>


int main()
{
char front[254];
int adapter = 0;
int s_adapter = 0;
int f_open;
int ioctl;
int request = FE_GET_INFO;
struct dvb_frontend_info info;


for (adapter = 0; adapter < 4; ++adapter)
{
printf("Adapter = %i\n", adapter);
for (s_adapter = 0; s_adapter != -1; ++s_adapter)
{
sprintf(front,"/dev/dvb/adapter%i/frontend%i", adapter, s_adapter);
printf("Prüfe Adapter: %s\n", front);
if((f_open = open(front,O_RDONLY)) < 0)
{
printf("Device '%s' konnte nicht geöfnnet werden\n", front);
s_adapter = -2;
}
else
{
if ( ioctl(f_open, request, info ) ) perror("ioctl");
printf("Status %s\n",info.name);
}
close(f_open);
}
}
}

-----------------
I compile this code very simple with following command:
gcc -o dvb_test2 dvb_test2.c

The error code I get is:
vb_test2.c:33: error: called object ‘ioctl’ is not a function

Is there anybody who can help me?
Thanks a lot!!!!

[johnl]

#include <stdio.h>
#include <fcntl.h>
#include <sys/ioctl.h>
#include <linux/dvb/dmx.h>
#include <linux/dvb/frontend.h>
#include <unistd.h> /* for close() */

int main()
{
char front[254];
int adapter = 0;
int s_adapter = 0;
int f_open;
/* you declared an unused(?) variable named ioctl which conflicts with
* the function named ioctl() you tried to use below. remove or rename
* this variable */
/* int ioctl; */
int request = FE_GET_INFO;
struct dvb_frontend_info info;


for (adapter = 0; adapter < 4; ++adapter)
{
printf("Adapter = %i\n", adapter);
for (s_adapter = 0; s_adapter != -1; ++s_adapter)
{
sprintf(front,"/dev/dvb/adapter%i/frontend%i", adapter, s_adapter);
printf("Prüfe Adapter: %s\n", front);
if((f_open = open(front,O_RDONLY)) < 0)
{
printf("Device '%s' konnte nicht geöfnnet werden\n", front);
s_adapter = -2;
}
else
{
if ( ioctl(f_open, request, info ) ) perror("ioctl");
printf("Status %s\n",info.name);
}
close(f_open);
}
}
}

[BGraaf]

Thanks a lot.

that something we called in Germany "Den Wald vor lauter Bäumen nicht sehen".
Now its working!

[Tony Flury]

Thanks a lot.

that something we called in Germany "Den Wald vor lauter Bäumen nicht sehen".
Now its working!

An equivalent in English is "You can't see the wood for the trees"