Why use (void)1 as a no-op in C++?

Question

I'm reviewing a third party codebase and see this definition of an assert macro:

#define assert( x ) \
     if( !( x ) ) { \
        ThrowException( __FILE__, __LINE__ ); \
     } else \
        ((void)1)

What's the point in (void)1? How is it better than idiomatic (void)0?

Just a lack of imagination. Replace with (void)42 so its obvious. — Hans Passant, Nov 02 '11 at 11:08
This is special: http://www.google.com/codesearch#nJHaZQ1IJ84/trunk/third_party/spidermonkey/js/src/jsdhash.c&q=%22%28void%291%22&type=cs, it uses `(void)1` and `(void)0`. Perhaps to distinguish between the two when examining preprocessed output? — Steve Jessop, Nov 02 '11 at 12:21

score 4 · Accepted Answer · answered Nov 02 '11 at 11:01

4

There's no difference between (void)1 and (void)0.

answered Nov 02 '11 at 11:01

Dietrich Epp

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@iammilind: The question was "How is [`(void)1`] better than idiomatic `(void)0`?" I answered it. It's not, there's no difference. – Dietrich Epp Nov 02 '11 at 11:24
@iammilind: I'm not sure there's is any insight to give. The asker already knows they're no-ops. – Dietrich Epp Nov 02 '11 at 11:49

score 4 · Answer 2 · answered Nov 02 '11 at 11:02

4

I think it does not matter that much (and will be optimized away by the compiler). And <cassert> is a standard C++ header (using the standard <assert.h> C header) which defines a standard assert macro, so an application should not re-define it.

answered Nov 02 '11 at 11:02

Basile Starynkevitch

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Why use (void)1 as a no-op in C++?

2 Answers2