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up vote 1 down vote favorite

I am reading a book to learn C. In that book is the following example code giving a preprocessor error with gcc (Debian 4.7.2-4) 4.7.2. The error is

file.c: In function ‘main’:
file.c:16:14: error: token ""I know the C language.\n"" is not valid in preprocessor expressions
file.c:20:14: error: token ""I know BASIC.\n"" is not valid in preprocessor expressions

The code is:

#include <stdio.h>

#define C_LANG    'C'
#define B_LANG    'B'
#define NO_ERROR  0

int main(void)
{
   #if C_LANG == 'C' && B_LANG == 'B'
     #undef C_LANG
     #define C_LANG "I know the C language.\n"
     #undef B_LANG
     #define B_LANG "I know BASIC.\n"
     printf("%s%s", C_LANG, B_LANG);
   #elif C_LANG == 'C'
     #undef C_LANG
     #define C_LANG "I only know C language.\n"
     printf("%s", C_LANG);
   #elif B_LANG == 'B'
     #undef B_LANG
     #define B_LANG "I only know BASIC.\n"
     printf("%s", B_LANG);
   #else
     printf("I don't know C or BASIC.\n");
   #endif

   return NO_ERROR;
}

Is the gcc preprocessor incapable of doing this correctly or is the something wrong with the code that needs to be changed?

share|improve this question
    
I'm thinking it's the printfs that are messing things up. –  Marvo May 20 '13 at 18:31
1  
No, it's the #elif lines that are having problems. For instance, the preprocessor is expanding the first #elif to #elif "I know the C language.\n" == 'C' and then failing on that. ideone fails it too. –  cebarth May 20 '13 at 18:39
    
Either of the answers given so far will work. However, I must say, if this is an example from a book, you may want to reconsider what book you are using. –  cebarth May 20 '13 at 18:58
    
@cebarth: I think you are right. The book is Sam's Teaching yourself C in 24 hours. It was a great for bash scripting and got me well on my way years ago with that. This really helped me grasp pointers when the Apress book C: Novice to Professional had me stumped on them, but it is using Windows with an unknown compiler. Perhaps the preprocessor is "smarter" which makes me sad for gcc. This is the 23rd chapter and thus 1 more to go. I'll finish it, but will definately be reading other books, too. –  narnie May 20 '13 at 19:08
1  
@narnie I-am-not-a-C-guru but I would suggest that such preprocessor code isn't really common/desirable. I think it's strange that gcc seems to evaluate the conditional expression in the elif even when the previous if has been checked to be non-zero, but I would maybe suggest that this sort of edge case isn't important. I think if you ever wanted code like this you'd at least want to use two different macro definitions like BLUEPIXY suggests. Also, I think if the book stated clearly at the beginning that it was using a different compiler, that you can't really fault it for that. –  roliu May 20 '13 at 19:52
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2 Answers

up vote 2 down vote

As @cebarth points out, the problem is that after you redefined C_LANG and B_LANG in the first #if, the #elif clauses fail, because the expansion is:

   #elif "I know the C language.\n" == 'C'
   /*...*/
   #elif "I know BASIC.\n" == 'B'

The C Standard says this about #if and #elif (C99 6.10.1):

Preprocessing directives of the forms
# if constant-expression new-line groupopt
# elif constant-expression new-line groupopt
check whether the controlling constant expression evaluates to nonzero.

There is no mention of not evaluating the expression because of an earlier check having succeeded.

One way to fix this is to redefine them back after the printf().

     #undef C_LANG
     #define C_LANG "I know the C language.\n"
     #undef B_LANG
     #define B_LANG "I know BASIC.\n"
     printf("%s%s", C_LANG, B_LANG);
     #undef C_LANG
     #define C_LANG 'C'
     #undef B_LANG
     #define B_LANG 'B'

Another way to fix this is to explicitly use #else instead of #elif.

   #if C_LANG == 'C' && B_LANG == 'B'
     #undef C_LANG
     #define C_LANG "I know the C language.\n"
     #undef B_LANG
     #define B_LANG "I know BASIC.\n"
     fprintf(stdout, "%s%s", C_LANG, B_LANG);
   #else
     #if C_LANG == 'C'
       #undef C_LANG
       #define C_LANG "I only know C language.\n"
       printf("%s", C_LANG);
     #elif B_LANG == 'B'
       #undef B_LANG
       #define B_LANG "I only know BASIC.\n"
       printf("%s", B_LANG);
     #else
       printf("I don't know C or BASIC.\n");
     #endif
   #endif
share|improve this answer
2  
Isn't the real question: Why does the preprocessor execute the elif code after the if succeeds? –  roliu May 20 '13 at 18:49
    
I don't think it is executing it, but rather just parsing the line. Does seem a bit silly though. –  cebarth May 20 '13 at 18:57
    
@roliu: The C Standard says only that the #if and #elif expressions are evaluated, and doesn't give any exceptions just because some prior #if or #elif has already matched. I will update the answer. –  jxh May 20 '13 at 19:26
    
"There is no mention of not evaluating the expression because of an earlier check having succeeded." - True, but GCC does mention it. "Then the text after each '#elif' is processed only if the '#elif' condition succeeds after the original '#if' and all previous '#elif' directives within it have failed" –  Mike May 20 '13 at 19:37
1  
@Mike, I think the "text after" refers to the conditional block, not the expression itself (i.e. 'groupopt', not 'constant-expression' in C99 6.10.1 item 3) –  cebarth May 20 '13 at 19:44
add comment
up vote 0 down vote 
#include <stdio.h>

#define C_LANG    'C'
#define B_LANG    'B'
#define NO_ERROR  0

int main(void)
{
   #if C_LANG == 'C' && B_LANG == 'B'
     #define C_LANG_VALUE "I know the C language.\n"
     #define B_LANG_VALUE "I know BASIC.\n"
     printf("%s%s", C_LANG_VALUE, B_LANG_VALUE);
   #elif C_LANG == 'C'
     #define C_LANG_VALUE "I only know C language.\n"
     printf("%s", C_LANG_VALUE);
   #elif B_LANG == 'B'
     #define B_LANG_VALUE "I only know BASIC.\n"
     printf("%s", B_LANG_VALUE);
   #else
     printf("I don't know C or BASIC.\n");
   #endif

   return NO_ERROR;
}
share|improve this answer
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