default value for macro declaration

Question

When a macro is created but no value is assigned, is there a default value? #define MACRO (end of line).

Macro definition has the syntax: #define <IDENTIFIER> <value>. Even if we don't pass a value, the macro is created, which can be verified by the following the program:

#include <stdio.h>
#define MACRO
int main() {
    #ifdef MACRO
        printf("MACRO was created");
    #endif
    
    return 0;
}

Also, if an identifier was not declared using #define but passed as the condition for #ifdef <IDENTIFIER>, the compiler treats the identifier as though it were the constant zero (false), which can be verified by the following the program:

#include <stdio.h>
int main() 
{
    #ifdef A
        printf("No print but valid condition");
    #endif
    
    return 0;
}

Which doesn't print anything, but there are no warnings nor errors from the compiler.

My question is, when a macro is created but no value is assigned, is there a default value?

Consider the program:

#include <stdio.h>
#define MACRO

int main() 
{
    #if defined MACRO
        printf("%d", MACRO);
    #endif
    
    return 0;
}

get the error message:

error: expected expression before ‘)’ token

Which would compile if we assign a value:

#include <stdio.h>
#define MACRO 3

int main() 
{
    #ifdef MACRO
        printf("%d", MACRO);
    #endif
    
    return 0;
}

My guess is that #define MACRO doesn't assign any value, even though #ifdef will treat not-defined macros as 0 (or a false condition); Is that correct?

KamilCuk · Accepted Answer · 2023-04-02T22:35:57.840

when a macro is created but no value is assigned, is there a default value?

No, or yes? A macro defined without a replacement list like so:

#define MACRO

expands to nothing, is removed from the processed input. Or the default value is nothing, an empty string, if you want to look at it that way.

#ifdef , the compiler treats the identifier as though it were the constant zero

#ifdef will treat not-defined macros as 0

No. #ifdef only cares about if defined or not defined. It doesn't care about the result of the expansion.

#if and #elif expand the expression and treat undefined(!) tokens as 0.

In other words, #if and #elif first expand all the macros, and then if there are any tokens that are not numbers or arithmetic operations, then those leftover tokens are replaced by 0.

You may want to read C11 standard 6.10.1p4 for the source, maybe it will clear things up:

[...] After all replacements due to macro expansion and the defined unary operator have been performed, all remaining identifiers (including those lexically identical to keywords) are replaced with the pp-number 0, and then each preprocessing token is converted into a token. The resulting tokens compose the controlling constant expression which is evaluated according to the rules ... [...]

My guess is that #define MACRO doesn't assign any value ... Is that correct?

 #define MACRO
 printf("%d", MACRO);

expands to printf("%d", );. The part , ); is invalid, thus you get a syntax error.

`#if FOO`\n`printf("Hello World!");`\n`#endif` will terminate with no errors and no warnings, here the `#if` directive, treats `FOO` as 0 (false), and thus, the condition `printf` will not run. Why is that if `FOO` is an "empty string"? — Darvid, Apr 02 '23 at 22:19
I do not understand. If `FOO` is _not_ defined, _then_ `#if FOO` is expanded to `#if 0`. If `FOO` is _defined_ to empty, _then_ `#if FOO` expands to `#if` and that is a compile error. Is FOO defined or not in your code? Consider posting a link to full code in godbolt.com. — KamilCuk, Apr 02 '23 at 22:32

default value for macro declaration

1 Answers1