问题
On my implementation of C++ (Visual Studio 2008 implementation) I see the following line in <cassert>
#ifdef NDEBUG
#define assert(_Expression) ((void)0)
I do not understand the need to cast 0 to void. It seems to me that
#ifdef NDEBUG
#define assert(_Expression) (0)
or even simply
#ifdef NDEBUG
#define assert(_Expression) 0
would do, considering the contexts in which assert(expr) can be used.
So, what's the danger of 0 of type int instead of 0 of type void in this case? Any realistic examples?
回答1:
The only purpose of the complicated expression (void)0 is to avoid compiler warnings. If you just had a naked, useless expression, the compiler might warn about an expression that has no effect. But by explicitly casting something to void you indicate that you mean to do this.
(Think about how confusing it would be to the user if the compiler suddenly said, "Warning: expression 0; has no effect.", when all you've done is switched to release mode.)
This was also common practice in C, where you'd say (void)printf("Hello"); to tell the compiler that you intentionally chose to ignore the return value of the function.
The (void) cast is not merely a choice by a particular implementation; it's required by the C standard. Quoting the 2011 ISO C standard (similar wording appears in the 1990 and 1999 editions):
If
NDEBUGis defined as a macro name at the point in the source file where<assert.h>is included, theassertmacro is defined simply as
#define assert(ignore) ((void)0)
The C++ standard requires the contents of the <cassert> header to be the same as the Standard C <assert.h> header.
来源：https://stackoverflow.com/questions/7002719/casting-0-to-void