• "Gisle Vanem" <giva bgnett.no> Dec 15 2002
• Richard <fractal clark.net> Dec 15 2002
• "Gisle Vanem" <giva bgnett.no> Dec 15 2002
• Richard <fractal clark.net> Dec 15 2002
• Richard <fractal clark.net> Dec 15 2002

The following code doesn't compile:

--------------------------------------
#include <stdio.h>
typedef unsigned char eth_address[6];
extern int set_addr (const eth_address *addr);

int main (void)
{
  return (0);
}
--------------------------------
Error is:
  test.c(3) : Error: type qualifiers and static can only appear in outermost
array
  of function parameter

What is this? This used to be no problem.

Gisle V.

# rm /bin/laden
/bin/laden: Not found

In article <ati7lr$26qm$1 digitaldaemon.com>, Gisle Vanem says...
The following code doesn't compile:

--------------------------------------
#include <stdio.h>
typedef unsigned char eth_address[6];
extern int set_addr (const eth_address *addr);

int main (void)
{
  return (0);
}


Not sure why it stopped working. I'm almost conviced that the fact that it does
not work is correct. The array [6] gets converted to an unsigned char * to array
element [0], then the parameter gets further defined as a pointer to a pointer
to unsigned char. Now the compiler should try and make the inner array constant,
but there is a rule to prevent that so you get the error. Try -

typedef unsigned char eth_address[6];
extern int set_addr (eth_address* const addr);
//extern int set_addr (const eth_address *addr);

int main (void)
{
return (0);
}

This marks the exterior "array" as constant.

I can't help wonder why your are passing a pointer to a pointer to unsigned
char. The array get's converted to a pointer to unsigned char automagically with
the following. Perhaps it is an external lib..

typedef unsigned char eth_address[6];
extern int set_addr (const eth_address addr);

..

// inside set_add
unsigned char* part = addr;
unsigned char* piece = addr[4];

..

Richard

"Richard" <fractal clark.net> wrote:

 In article <ati7lr$26qm$1 digitaldaemon.com>, Gisle Vanem says...
The following code doesn't compile:

--------------------------------------
#include <stdio.h>
typedef unsigned char eth_address[6];
extern int set_addr (const eth_address *addr);

int main (void)
{
  return (0);
}


 Not sure why it stopped working. I'm almost conviced that the fact that it does
 not work is correct.


It works with other compilers I've tried (bcc 4.5, gnu C 3.21, Metaware HighC
3.61,
Watcom C 11 and MS Quick-C v8

The array [6] gets converted to an unsigned char * to array
 element [0], then the parameter gets further defined as a pointer to a pointer
 to unsigned char. Now the compiler should try and make the inner array
constant,
 but there is a rule to prevent that so you get the error. Try -


 typedef unsigned char eth_address[6];
 extern int set_addr (eth_address* const addr);


That worked, but is it the same? I want the '*addr' to
be const, not the 'addr' itself.

Gisle V.

In article <atijoo$2iam$1 digitaldaemon.com>, Gisle Vanem says...

 typedef unsigned char eth_address[6];
 extern int set_addr (eth_address* const addr);


That worked, but is it the same? I want the '*addr' to
be const, not the 'addr' itself.


Expanding the parameter out you get

unsigned char * * const addr

Which reads: addr is an ordinary pointer that is const that points to an
ordinary pointer that points to unsigned char. So yes, addr is const and its
value cannot be changed, but what it points to can be changed.

If you want to qualify what the array points to as const (the 6 unsigned char
can't be changed) you can try:

typedef const unsigned char * eth_address;
extern int set_addr (eth_address* addr);

which reads: addr is an ordinary pointer that points to an ordinary pointer that
points to const unsigned char.

Richard

This approach might suit your needs as well, but you need to lose the typedef.

extern int set_add(const unsigned char (*addr)[6]);

int set_addr(const unsigned char (*addr)[6])
{
const unsigned char *a = *addr;
return 0;
}

void main()
{
unsigned char b[6] = "12345";
set_addr(&b);

}