C preprocessor

[ee02iu]

I was told C preprocessor "#define arg1 agr2" would do a text replacement - arg1 string with arg2....

I wrote a few lines of code as follows:

#include "stdio.h"
#define PRNT(v,x) printf("x = %v\n", x)

main() {
int i = 5;
PRNT(d, i);
return 0;
}

I expected that after the preprocessing step, the program would become

main() {
int i = 5;
printf("i = %d\n", i);
return 0;
}

and it should print
i = 5

but the preprocessor #define did not work as expected! - the program printed

i = %v

???? Why does C behave that way? What didn't I understand? Please help
Thanks
EE02IU
P.S. I used Borland C++ 4.52

[Banfa]

I was told C preprocessor "#define arg1 agr2" would do a text replacement - arg1 string with arg2....

I wrote a few lines of code as follows:

#include "stdio.h"
#define PRNT(v,x) printf("x = %v\n", x)

main() {
int i = 5;
PRNT(d, i);
return 0;
}

a #defined symbol (PRNT in your case) is used for a text replacement...

except inside strings, the contents of strings are not parsed by the pre-processor for the replacement of #defined symbols

However you can achieve what you want like this

#define PRNT(v,x)   printf("x = %" #v "\n", x)
 

This makes use of the following 2 features onf the C language

1. Separate strings seperated only by white space are concatinated by the C compiler, for instance

  printf("H"
      "ell"
      "o Wor"
      "ld\n");
 

outputs

Hello World

This is useful if you have very long strings as you don't have to fit them on one line, or for laying out multiline messages as you can put different output lines on different source lines

2. The preprocessor stringizing token #. #v is iterpreted by the preprocessor as a string containing whatever is in the parameter v