What does void casting do?

Question

I was reading some source code, and this came up;

struct Cookie *
Curl_cookie_add(struct SessionHandle *data, /* rest of params were here */)
{
/* unrelated things were here */
#ifdef CURL_DISABLE_VERBOSE_STRINGS
  (void)data;
#endif
/* rest of function goes here */
}

As you can see, void casted pointer, isn't even assigned to a variable. I was wondering what is the purpose of this.

Is this a raw copy-pasta? is it possible that its casted to a `void *`, not void? — Richard J. Ross III, May 26 '12 at 16:23
possible duplicate of [casting unused return values to void](http://stackoverflow.com/questions/689677/casting-unused-return-values-to-void) — Adam Liss, May 26 '12 at 16:24
Please search the existing questions and answers before posting. — Adam Liss, May 26 '12 at 16:24
@RichardJ.RossIII yes, this is raw copy paste; check this out: https://github.com/bagder/curl/blob/master/lib/cookie.c#L198 — yasar, May 26 '12 at 16:25
@AdamLiss nope, not a dupe. this is about a struct, not a function. — Richard J. Ross III, May 26 '12 at 16:25
Not true. In both cases an unused value is cast to `void` to avoid a compiler warning. This one happens to be a pointer to a structure, but it's still an unused function argument. — Adam Liss, May 26 '12 at 16:26

score 6 · Accepted Answer · answered May 26 '12 at 16:29

6

This cast suppresses a compiler warning that would arise if data is not used.

GCC produces this warning if the -Wunused-parameter flag (implied by -Wextra) is enabled.

answered May 26 '12 at 16:29

Joey Adams

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What compiler/flags would be needed for that warning? With GCC, I have no issues w/ that. – Richard J. Ross III May 26 '12 at 16:29
@RichardJ.RossIII: It appears with `-Wall -W`, but I'm still looking for the exact flag. – Joey Adams May 26 '12 at 16:31
This answer doesn't answer the question *Why* - it gives a method of hiding the problem? I suggest that suppressing warnings is not an ideal solution. – Andrew Aug 16 '15 at 08:36
@Andrew: This is selective suppression of a warning, so the coder can indicate that a particular function purposely does not use a particular parameter. When you don't use a variable, it's often a mistake or unfinished code, except in some cases (e.g. a callback function where parameters are provided in case you need them). So we want unused warnings except in those special cases. – Joey Adams Aug 17 '15 at 16:54

score 0 · Answer 2 · answered Aug 18 '15 at 13:27

0

Fair point Joey - but suppressing that warning also suppresses it for all the cases where the programmer has accidentally not used a parameter...

answered Aug 18 '15 at 13:27

Andrew

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What does void casting do?

2 Answers2

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