passes of the Preprocessor //

[onkar]

#include<stdio.h>
#define string(x) #x
#define replace(x) string(x)
#define var Hello
int main(int argc,char **argv){
char *str=replace(var);
printf("%s\n",str);
return 0;
}

This will print :
Hello

Ok fine , but can any one explain me how the preprocessor does it ?? In
one pass on multiple passes .The answer will be invaluabe to me.

regards,
Onkar

 

[Keith Thompson]

#include<stdio.h>
#define string(x) #x
#define replace(x) string(x)
#define var Hello
int main(int argc,char **argv){
char *str=replace(var);
printf("%s\n",str);
return 0;
}

This will print :
Hello

Ok fine , but can any one explain me how the preprocessor does it ?? In
one pass on multiple passes .The answer will be invaluabe to me.

As far as I know, most preprocessors work in one pass, but why does it
matter? It works as defined by the standard (if it's not buggy); what
more do you really need to know?

(There's certainly nothing wrong with curiosity, but you said the
answer will be invaluable.)

 

[raxitsheth2000]

if you are using gcc, then there is -E option that will only Stop after
Pre-Comilation.
after that you can view the output file.


--raxit

 

[Richard Heathfield]

onkar said:

#include<stdio.h>
#define string(x) #x
#define replace(x) string(x)
#define var Hello
int main(int argc,char **argv){
char *str=replace(var);
printf("%s\n",str);
return 0;
}

This will print :
Hello

Ok fine , but can any one explain me how the preprocessor does it ?? In
one pass on multiple passes .The answer will be invaluabe to me.

http://c-faq.com Answer 11:17 explains this.

 

[CBFalconer]

if you are using gcc, then there is -E option that will only Stop
after Pre-Comilation. after that you can view the output file.

Don't top-post. See the following links.

--
Some informative links:
< <http://www.geocities.com/nnqweb/>
<http://www.catb.org/~esr/faqs/smart-questions.html>
<http://www.caliburn.nl/topposting.html>
<http://www.netmeister.org/news/learn2quote.html>
<http://cfaj.freeshell.org/google/>