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I started to look into glibc (GNU Libc) to understand how it's written. In malloc.c, I found a piece of code as follow:

#ifndef void
#define void        void
#endif

Can someone please explain to me what this means? Isn't void always defined?

Thanks

Orion
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4 Answers4

10

Looking at the git history, it was like this:

/*
  Void_t* is the pointer type that malloc should say it returns
*/

#ifndef Void_t
#if (__STD_C || defined(WIN32))
#define Void_t      void
#else
#define Void_t      char
#endif
#endif /*Void_t*/

This was a workaround for historical [C], which did not have void and malloc() returned char * instead of void. The code was removed by Ulrich Drepper in 2011. The commit does not seems to be generated by a script or anything automatic, so he must had his intension to define like that.

The commit message does not say anything about void:

Simplify malloc code

Remove all kinds of unused configuration options and dead code.

Yasushi Shoji
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    Thanks for the link. It was very helpful. I saw the removal of the above code and all the changes from `Void_t` to `void`, but `#define void void` still doesn't make sense to me. – Orion Aug 16 '12 at 03:42
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    It doesn't to me, neither. I've checked glibc code to see if there is `#ifdef void` to do a conditional compilation, but I couldn't find any. It might be better to just ask him or on the glibc mailing list. But I don't know. – Yasushi Shoji Aug 16 '12 at 03:47
  • I’m extremely certain this is just a botched part of the refactoring and can just be removed. – mirabilos Nov 25 '21 at 00:40
8

I don't know for certain what the reason for the #define void void is in malloc.c, but my guess follows:

As Yasushi Shoji mentioned, void was not always a keyword in C. When it was introduced/standardized, a common workaround to be able to compile code using the new void keyword with compilers that didn't support it was to define void as a macro such as:

#define void int   /* or maybe #define void char */

that macro definition might be done using the compiler command line instead of via a header so that there wouldn't be a need to ensure that all the translation units included a header that defined the macro.

However, it might also be common for programmers migrating to the new keyword to use sequences of code that looked like:

#ifndef void
#define void int
#endif

for example, you'll see the following bit of code: 

/*
 * This is a fairly bogus thing to do, but there seems to be no better way for
 * compilers that don't understand void pointers.
 */
#ifndef void
#define void char
#endif

in http://buildbot.zope.org/Squid-2.4STABLE6%2B/include/snmp_impl.h?annotate=1.1.1.1&cvsroot=Zope.org

So, my guess is that the #define void void in malloc.c is just a way to prevent any such sequences that might exist in the headers it includes from redefining void, yet still allow it to have been previously defined 'gloablly' (there are only comments in malloc.c before the #define void void) in case it was being compiled on a configuration that didn't support void. In other words, if void wasn't globally defined as a macro before when malloc.c was started to be compiled, there was no reason something should define to be something else later on in the compile.

Community
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Michael Burr
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4

Although void is a keyword in C, keywords are not defined as preprocessor symbols. The code you quoted ensures that it is also defined as a preprocessor symbol.

I don't know why this would be necessary.

Greg Hewgill
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0

void may always have a meaning, but I'd say void isn't normally #defined. I don't know exactly what's going on there, but the consequence of this run of code is that if anyone says #ifdef void later on, it'll be true. So in other words, it'll only get #defined to void once by this run of code.

sblom
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