CODEWITHSUNDEEP
c-preprocessorstringification
Alex Morales
Alex Morales

Reputation: 2791

C Preprocessor, Stringify the result of a macro

I want to stringify the result of a macro expansion.

I've tried with the following:

#define QUOTE(str) #str
#define TEST thisisatest
#define TESTE QUOTE(TEST)

And TESTE gets expanded to: "TEST", while I'm trying to get "thisisatest". I know this is the correct behavior of the preprocessor but can anyone help me with a way to achieve the other one?

Using TESTE #TEST is not valid
Using TESTE QUOTE(thisisatest) is not what I'm trying to do

Upvotes: 52

Views: 31240

Answers (2)

pratikm
pratikm

Reputation: 4254

To clarify a bit more, essentially the preprocessor was made to execute another "stage". i.e :

1st case:

->TESTE
->QUOTE(TEST) # preprocessor encounters QUOTE 
 # first so it expands it *without expanding its argument* 
 # as the '#' symbol is used
->TEST

2nd case:

->TESTE
->EXPAND_AND_QUOTE(TEST)
->QUOTE(thisisatest) 
  # after expanding EXPAND_AND_QUOTE
  # in the previous line
  # the preprocessor checked for more macros
  # to expand, it found TEST and expanded it
  # to 'thisisatest'
->thisisatest

Upvotes: 18

Steve Jessop
Steve Jessop

Reputation: 279195

Like this:

#include <stdio.h>

#define QUOTE(str) #str
#define EXPAND_AND_QUOTE(str) QUOTE(str)
#define TEST thisisatest
#define TESTE EXPAND_AND_QUOTE(TEST)

int main() {
    printf(TESTE);
}

The reason is that when macro arguments are substituted into the macro body, they are expanded unless they appear with the # or ## preprocessor operators in that macro. So, str (with value TEST in your code) isn't expanded in QUOTE, but it is expanded in EXPAND_AND_QUOTE.

Upvotes: 85

Related Questions