# and ## operators

[Mantorok Redgormor]

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.


nethlek

[Emmanuel Delahaye]

In 'comp.lang.c', ne*****@tokyo.com (Mantorok Redgormor) wrote:

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.

I use it every day. For example, it very useful for local debug macros.

Say, I have a structure that represent some configuration data:

typedef struct
{
unsigned int foo;
some_enum_type bar;
unsigned long baz;
}
config_s;

At a certain stage of my code, I want a trace of the values on the console.

Ok, I can write

printf ("foo = %u", p->foo);
printf ("bar = %u", (unsigned int) p->bar);
printf ("baz = %ul", p->baz);

which is rapidly boring is I have 20 elements or more...

What can be done to help and avoid errors is to ask the proprocessor to
write the code for you:

#define PRT(a) \
printf ("%4s = %ul", #baz, (unsigned long) p-> ## a)

PRT (foo);
PRT (bar);
PRT (baz);

#undef PRT

--
-ed- em**********@noos.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
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[Kevin Easton]

Emmanuel Delahaye <em**********@noos.fr> wrote:

In 'comp.lang.c', ne*****@tokyo.com (Mantorok Redgormor) wrote:

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.

I use it every day. For example, it very useful for local debug macros.

Say, I have a structure that represent some configuration data:

typedef struct
{
unsigned int foo;
some_enum_type bar;
unsigned long baz;
}
config_s;

At a certain stage of my code, I want a trace of the values on the console.

Ok, I can write

printf ("foo = %u", p->foo);
printf ("bar = %u", (unsigned int) p->bar);
printf ("baz = %ul", p->baz);

which is rapidly boring is I have 20 elements or more...

What can be done to help and avoid errors is to ask the proprocessor to
write the code for you:

#define PRT(a) \
printf ("%4s = %ul", #baz, (unsigned long) p-> ## a)

I think you mean #a there. In the expression p->bar, there are three
tokens: "p", "->" and "bar". So you don't need the ## token pasting
operator here - this definition works:

#define PRT(a) \
printf ("%4s = %ul", #a, (unsigned long) p-> a)

To the OP: the ## operator is needed whenever you need to create a
single token from multiple tokens - this only ever occurs in macros.
Say you had some code like this:

func_a(s->data_a);
func_b(s->data_b);
func_c(s->data_c);
func_d(s->data_d);
func_e(s->data_e);

and you wanted to use a macro to make it less error-prone. You could do
that like this:

#define DO_THING(x) func_ ## x (s -> data ## x)

DO_THING(a);
DO_THING(b);
DO_THING(c);
DO_THING(d);
DO_THING(e);

- Kevin.

[Emmanuel Delahaye]

In 'comp.lang.c', Kevin Easton <[e-mail]> wrote:

#define PRT(a) \
printf ("%4s = %ul", #baz, (unsigned long) p-> ## a)
I think you mean #a there.

Oops, absolultly. Yet another victim of copy and paste...
In the expression p->bar, there are three
tokens: "p", "->" and "bar". So you don't need the ## token pasting
operator here - this definition works:
True enough, my example was poor.
#define PRT(a) \
printf ("%4s = %ul", #a, (unsigned long) p-> a)

--
-ed- em**********@noos.fr [remove YOURBRA before answering me]
The C-language FAQ: http://www.eskimo.com/~scs/C-faq/top.html
<blank line>
FAQ de f.c.l.c : http://www.isty-info.uvsq.fr/~rumeau/fclc/

[mbs]

> When is it appropriate to use # and ##?

the # operator helps you harness the power of the c-preprocessor.

suppose you define a macro like this:
#define SHOW_INT( var ) printf( #var"=%d\n", var );

and call it later:

int i=800;
...
SHOW_INT( i );
...

it would expend to
printf( "i""=%d\n", i );
and would print
i=800
to stdout.

Here is my version of SHOW, which I use daily:
#define SHOW( val, fmt ) { fprintf( stderr, "%d "#val"=#fmt"\n",
__LINE__, (val) ); }

this macro prints the what it is about to do to stderr and the does
it:
#define DODBG( cmd ) { fprintf( stderr, "%d doing "#cmd"\n",
__LINE__ ); {cmd;} }
..

I have been told that ## is not useful at all

It is. There's an example in K&R, but the book is not here now.

Michael.

[Jack Klein]

On 20 Sep 2003 22:53:14 -0700, ne*****@tokyo.com (Mantorok Redgormor)
wrote in comp.lang.c:

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.

On the other hand, my dental hygienist told me just last week that ##
is very useful.

Perhaps the person who told you this is not useful at all.

--
Jack Klein
Home: http://JK-Technology.Com
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[Steve Zimmerman]

Mantorok Redgormor wrote:

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.

Mantorok,

Thank you for your post. I don't know if ## is useful at all.
The following program uses ##, but not in a very useful way.
#include <stdio.h>
#include <string.h>

#define paste(front, back) front ## back /* p. 92, K&R2 */

int main()
{
char *paste(Rama, nujan); /* char *Ramanujan; */

strcpy(Ramanujan, "Ramanujan is great");

printf("%s\n", paste(Rama, nujan));

return 0;
}

Output of program: Ramanujan is great
Aborted

Why does this program's output contain two streams (stdout and stderr)?

--Steve

[Irrwahn Grausewitz]

Steve Zimmerman <st******@sonic.net> wrote:

Mantorok Redgormor wrote:

When is it appropriate to use # and ##? I have been told that ## is
not useful at all.

Mantorok,

Thank you for your post. I don't know if ## is useful at all.
The following program uses ##, but not in a very useful way.
#include <stdio.h>
#include <string.h>

#define paste(front, back) front ## back /* p. 92, K&R2 */

int main()
{
char *paste(Rama, nujan); /* char *Ramanujan; */

strcpy(Ramanujan, "Ramanujan is great");

You failed to allocate some memory for 'Ramanujan' to point to.

printf("%s\n", paste(Rama, nujan));

return 0;
}

Output of program: Ramanujan is great
Aborted

Why does this program's output contain two streams (stdout and stderr)?

The abort message, induced by the error above, was printed to stderr.

Regards

Irrwahn
--
My other computer is a abacus.