Reg: #define

[raghu]

#define GOOGLE
int main(void)
{
printf("%d",GOOGLE);
return 0;
}

In the above program ,by default GOOGLE should be assigned to
zero..right?
But when I try to print it it gives an error at printf
statement....Why?
If I remove the comma before the GOOGLE it compiles succesfully..Why
so?

Please explain..

Thank You.

Regards,
Raghu

[Walter Roberson]

In article <11*********************@n59g2000hsh.googlegroups. com>,
raghu <ra*********@gmail.comwrote:

>#define GOOGLE
int main(void)
{
printf("%d",GOOGLE);
return 0;
}

>In the above program ,by default GOOGLE should be assigned to
zero..right?

No. GOOGLE would only be substituted as 0 in evaluating #if
conditions. Otherwise, it is substituted as nothing at all.
That would make your printf statement into

printf("%d",)

which is a syntax error.
--
Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson

[Keith Thompson]

"raghu" <ra*********@gmail.comwrites:

#define GOOGLE
int main(void)
{
printf("%d",GOOGLE);
return 0;
}

In the above program ,by default GOOGLE should be assigned to
zero..right?
But when I try to print it it gives an error at printf
statement....Why?

Walter Roberson already answered that.

If I remove the comma before the GOOGLE it compiles succesfully..Why
so?

Without the comma, the line is equivalent to

printf("%d");

which is wrong, but it's not an error that the compiler is required to
detect. With a "%d" format, you need an argument of type int -- but
the compiler doesn't necessarily know that. (Some compilers are
clever enough to warn you about such things.) If you execute the
program, the behavior is undefined.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <* <http://users.sdsc.edu/~kst>
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[saki]

On Apr 11, 11:53 pm, [e-mail] (Walter Roberson)
wrote:

In article <[e-mail]. com>,

raghu <[e-mail]:

#define GOOGLE
int main(void)
{
printf("%d",GOOGLE);
return 0;
}
In the above program ,by default GOOGLE should be assigned to
zero..right?

No. GOOGLE would only be substituted as 0 in evaluating #if
conditions. Otherwise, it is substituted as nothing at all.
That would make your printf statement into

printf("%d",)

which is a syntax error.
--
Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson

could u just explain a little more as to how it happens

[Walter Roberson]

In article <11**********************@o5g2000hsb.googlegroups. com>,
saki <sa***********@gmail.comwrote:

>On Apr 11, 11:53 pm, [e-mail] (Walter Roberson)
wrote:

>In article <[e-mail]. com>,

>raghu <[e-mail]:

>#define GOOGLE
int main(void)
{
printf("%d",GOOGLE);
return 0;
}
In the above program ,by default GOOGLE should be assigned to
zero..right?

>No. GOOGLE would only be substituted as 0 in evaluating #if
conditions. Otherwise, it is substituted as nothing at all.

Correction: GOOGLE would be substituted as nothing at all
even in #if statements. The implicit substitution of 0 for
identifiers only holds for #if statements when the identifier
has *not* been #define'd .

>That would make your printf statement into
printf("%d",)
which is a syntax error.

>could u just explain a little more as to how it happens

Once you have #define'd something using an "object-like" definition
(that is, there is no parameter list on the #define), then
whereever that macro name occurs that is not in quotes, the
-exact- sequence of characters you #define'd it as will be put
into the line, just as if they had always been there. If you
didn't put in any replacement sequence then there will be nothing
to insert there, and the name will effectively be deleted, as if
you had not written it there at all.

[The above is not quite technically correct in certain cases
involving recursive preprocessing macros, and is not technically
correct in certain cases involving object macros defined using the
# operator, such as #define StringMe(foo) #foo ]

If you have #define'd something, then even if you didn't
specify any replacement text, then any #ifdef preprocessor statement
will know that the macro has been defined, and any use of
the defined() operator on an #if statement will know that the
macro has been defined. For example,

If something has -not- been #define'd, then in any #if statement,
it will be replaced by 0. That does not hold if you have #define'd
it in any way: if you #define'd it, the definition you gave will
be substituted in.

#define GOOGLE
#if defined(GOOGLE)
/* This part -will- be included */
#endif
#ifdef GOOGLE
/* This is the same as above; GOOGLE has been defined so
this part -will- be included */
#endif
#if defined(ROCKS)
/* but ROCKS has not been defined, so this part will -not- be included */
#endif
#if !ELGOOG
/* ELGOOG has not been defined, so it will be replaced by 0. You
then have !0 which is 1 which is true, so the #if will succeed
and this part -will- be included */
#endif
--
"law -- it's a commodity"
-- Andrew Ryan (The Globe and Mail, 2005/11/26)

[army1987]

On 11 Apr, 20:53, [e-mail] (Walter Roberson) wrote:

No. GOOGLE would only be substituted as 0 in evaluating #if
conditions. Otherwise, it is substituted as nothing at all.

No. It is replaced with whitespace.
i+GOOGLE+; becames i+ +; which is a syntax error, not i++;.

[Eric Sosman]

ar******@email.it wrote On 04/13/07 07:06,:

On 11 Apr, 20:53, [e-mail] (Walter Roberson) wrote:

>>No. GOOGLE would only be substituted as 0 in evaluating #if
conditions. Otherwise, it is substituted as nothing at all.

No. It is replaced with whitespace.
i+GOOGLE+; becames i+ +; which is a syntax error, not i++;.

No; it is replaced with nothing at all. The reason
`i+GOOGLE+;' is a syntax error is not because GOOGLE is
replaced by white space (it isn't), but because the input
has already been divided into the preprocessor tokens

i
+
GOOGLE
+
;

before the macro is recognized and replaced.

Describing the preprocessor's actions in terms of text
replacement is tempting and may even make them easier for
a new learner to understand, but the description is wrong
at a fundamental level. The preprocessor operates on the
preprocessor tokens that have already been formed from
source text, not on the source text characters themselves.
Once in a while, the distinction is important, as in

#define YAHOO E+6
double million = 1.YAHOO; /* syntax error */

--
Er*********@acm-dot-org.invalid