Preprocessor directive

[Trying_Harder]

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how?

/If/ my question above isn't very clear you can refer to
the following example.

eg cosider 2 files sample.c and sample.h

sample.h
#define BLAH 10

sample.c
/* This should print 10 */
printf("\n %d ", BLAH);

func1();

printf("\n %d ", BLAH);

func1()
{
#undef BLAH
#define BLAH 15
/* This should print 15, but BLAH has a local scope */
printf("\n %d ", BLAH);

}

Thanks,

[Kris Wempa]

What are you trying to accomplish ? The pre-processor simply expands your
macro when you compile your code. So, if you were to compile a separate .c
file that included 'sample.h', it would only see the definition of BLAH in
sample.h (which is 10). BLAH does not have local scope. As far as the
compiler is concerned, the argument to the last printf() statement is the
constant 15. Explain what you are trying to do and I can provide some
assistance.

"Trying_Har der" <fr***********@ yahoo.com> wrote in message
news:b0******** *************** ***@posting.goo gle.com...

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how?

/If/ my question above isn't very clear you can refer to
the following example.

eg cosider 2 files sample.c and sample.h

sample.h
#define BLAH 10

sample.c
/* This should print 10 */
printf("\n %d ", BLAH);

func1();

printf("\n %d ", BLAH);

func1()
{
#undef BLAH
#define BLAH 15
/* This should print 15, but BLAH has a local scope */
printf("\n %d ", BLAH);

[Artie Gold]

Trying_Harder wrote:

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how?
Macros are not `scoped'. Preprocessing takes place *before* scope of
any kind is established.
/If/ my question above isn't very clear you can refer to
the following example.

eg cosider 2 files sample.c and sample.h

sample.h
#define BLAH 10

sample.c
/* This should print 10 */
printf("\n %d ", BLAH);

func1();

printf("\n %d ", BLAH);

func1()
{
#undef BLAH
#define BLAH 15
/* This should print 15, but BLAH has a local scope */
printf("\n %d ", BLAH);

}

`BLAH' will be replaced by `15' for the rest of the translation unit.

HTH,
--ag

--
Artie Gold -- Austin, Texas
Oh, for the good old days of regular old SPAM.

[Eric Sosman]

Trying_Harder wrote:

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how?
The scope of a macro definition lasts from the #define
to the #undef, or to the end of the translation unit. Macro
scopes do not nest, so there's no way to "push" a replacement
definition and then "pop" the original.
/If/ my question above isn't very clear you can refer to
the following example.

eg cosider 2 files sample.c and sample.h

sample.h
#define BLAH 10

sample.c
/* This should print 10 */
printf("\n %d ", BLAH);

func1();

printf("\n %d ", BLAH);

func1()
{
#undef BLAH
#define BLAH 15
/* This should print 15, but BLAH has a local scope */
printf("\n %d ", BLAH);

}

The numbers output would be 10, 15, and 10, in that order.
BLAH does not have "a local scope" in func1(); BLAH remains
defined as 15 all the way to the end of the compilation.

--
Er*********@sun .com

[Charles Harrison Caudill]

Trying_Harder <fr***********@ yahoo.com> wrote:

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how? /* This should print 15, but BLAH has a local scope */

preprocessor directives don't have scope. The preprocessor will go from
the beginning of the program to the end and make replacements as necessary.
The first two BLAH's are being replaced by 10 because when test.h is
#included, it is #defined to 10. The BLAH in func1 was replaced by 15
because it was defined after main. If you put its declaration above main
then it will print 15 for all of them. Keep in mind, the preprocessor
is *not* operated at run-time, it is done *before* compilation.

What are you trying to accomplish? Even if you could pull that off you
probably shouldn't.

--
Harrison Caudill | .^ www.hypersphere.org
Computer Science & Physics Double Major | | Me*Me=1
Georgia Institute of Technology | '/ I'm just a normal guy

[Trying_Harder]

>

preprocessor directives don't have scope. The preprocessor will go from
the beginning of the program to the end and make replacements as necessary.
The first two BLAH's are being replaced by 10 because when test.h is
#included, it is #defined to 10. The BLAH in func1 was replaced by 15
because it was defined after main. If you put its declaration above main
then it will print 15 for all of them. Keep in mind, the preprocessor
is *not* operated at run-time, it is done *before* compilation.

What are you trying to accomplish? Even if you could pull that off you
probably shouldn't.
Let me start off with a thanks to all.
Actually, I have 3 modes the program can execute in, for example sake
`a', `b' and `c'. Now in each of these modes value of a particular
preprocessor macro is different, but the program can execute only in
a one mode in one execution.
Decision of what its value is, is taken soon after the program begins,
but I am doing this part in a function because of my aim to keep the
main function "light".

So func1() is where I am assigning the value to these macros depending
on `argv'.
One solution is to return a value from func1() hinting something about
the mode and #defining in main.
But, can't this be done in func1()?

Btw, in one of the other posts

<quoting Eric Sosman>

{
#undef BLAH
#define BLAH 15
/* This should print 15, but BLAH has a local scope */
printf("\n %d ", BLAH);

}

The numbers output would be 10, 15, and 10, in that order.
BLAH does not have "a local scope" in func1(); BLAH remains
defined as 15 all the way to the end of the compilation.

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^

If don't find this answer not convincing. If this was true then why
is the 3rd `printf' (in main) printing out 10 ? According to this
explanation it should print out 15 everywhere (after this "redefiniti on")
regardless of the scope.

Thank you for your help, I would really appreciate if someone could
refer me to some online resources containing preprocessor directives
explained in detail.

Thanks,

[dam_fool_2003]

Charles Harrison Caudill <ku*****@myrna. cc.gatech.edu> wrote in message news:<bl******* ***@solaria.cc. gatech.edu>...

Trying_Harder <fr***********@ yahoo.com> wrote:

Is it possible to redefine a macro with global scope after
undefining it in a function? If yes, could someone explain
how?

/* This should print 15, but BLAH has a local scope */

preprocessor directives don't have scope. The preprocessor will go from
the beginning of the program to the end and make replacements as necessary.
The first two BLAH's are being replaced by 10 because when test.h is
#included, it is #defined to 10. The BLAH in func1 was replaced by 15
because it was defined after main. If you put its declaration above main
then it will print 15 for all of them. Keep in mind, the preprocessor
is *not* operated at run-time, it is done *before* compilation.

^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^
What happens if macro is decleared in main().Just a doubt, that is all.
#define blaa 10
void fun(void)
{
printf("%d\n",b laa);
#ifdef blaa
#undef blaa
#define blaa 90
#endif
printf("now blaa is = %d\n",blaa);
}
int main(void)
{
fun();
#ifdef blaa
#undef blaa
#define blaa 98
#endif
printf("now blaa has %d\n",blaa);
return 0;
}

What are you trying to accomplish? Even if you could pull that off you
probably shouldn't.

[Trying_Harder]

fr***********@y ahoo.com (Trying_Harder) wrote in message > >

If don't find this answer not convincing. If this was true then why

^^^

Sorry about the typo there. I meant to say "I dont find this answer
convincing".

p.s: Can someone please attend to my question above?

Thanks,

[Jeremy Yallop]

Trying_Harder wrote:

p.s: Can someone please attend to my question above?

Here's the original code:

[Trying_Harder]
| sample.h
| #define BLAH 10
|
| sample.c
| /* This should print 10 */
| printf("\n %d ", BLAH);
|
| func1();
|
| printf("\n %d ", BLAH);
|
| func1()
| {
| #undef BLAH
| #define BLAH 15
| /* This should print 15, but BLAH has a local scope */
| printf("\n %d ", BLAH);
|
| }

Here's the (correct) answer you don't find convincing:

[Eric Sosman]
|> The numbers output would be 10, 15, and 10, in that order.
|> BLAH does not have "a local scope" in func1(); BLAH remains
|> defined as 15 all the way to the end of the compilation.

Here's the question I think you're referring to:

[Trying_Harder]
| If don't find this answer not convincing. If this was true then why
| is the 3rd `printf' (in main) printing out 10 ? According to this
| explanation it should print out 15 everywhere (after this "redefiniti on")
| regardless of the scope.

Preprocessor replacement works on the text of the source file, so
occurrences of "BLAH" in the lines following the "#undef", "#define"
directives are replaced with "15" during an early stage of
compilation. Occurrences of "BLAH" before this (but following the
first "#define") are replaced with "10". After all this has taken place
the output from "sample.c" looks something like this:

printf("\n %d ", 10);

func1();

printf("\n %d ", 10);

func1()
{

printf("\n %d ", 15);

}

Note that this all takes place at "compile-time", i.e. before the
program is run at all. Further stages of compilation translate your
program into some sort of executable format, by which point there is
no trace of "#define" or "BLAH" left; the preprocessor directives have
already been executed and only "runnable" code remains.

"BLAH" is not like a variable, which is initialised at runtime.
"func1()" does not change the value of "BLAH". The fact that the
second "#define" is textually inside the body of "func1()" is
irrelevant; there is nothing left of "BLAH" or its various definitions
by the time "func()" is called.

Jeremy.

[Jens.Toerring]

da***********@y ahoo.com wrote:

What happens if macro is decleared in main().Just a doubt, that is all.
Nothing special about that, the preprocessor does not care if it's in
main() or somewhere else:
#define blaa 10
void fun(void)
{
printf("%d\n",b laa);
This would be replaced by the preprocessor by

printf("%d\n",1 0);
#ifdef blaa
#undef blaa
#define blaa 90
#endif
Since 'blaa' was already defined it now would be redefined to 90
printf("now blaa is = %d\n",blaa);
so this would end up as

printf("now blaa is = %d\n",90);
}
int main(void)
{
fun();
fun will now print out

10
90

(BTW this would also happen if you would redefine 'blaa' before the
call of fun() since the preprocessor has already done it's work on
the body of fun().)
#ifdef blaa
#undef blaa
#define blaa 98
#endif
And this will redefine 'blaa' to 98
printf("now blaa has %d\n",blaa);
so here you get
printf("now blaa has %d\n",98); return 0;
}

Regards, Jens
--
_ _____ _____
| ||_ _||_ _| Je***********@p hysik.fu-berlin.de
_ | | | | | |
| |_| | | | | | http://www.physik.fu-berlin.de/~toerring
\___/ens|_|homs|_|oe rring