A question about #define

[George Ma]

Hi, everyone!

Please look at the following statement,

--------
#define assert(x) ((void)0)
--------

What does the statement mean? What is the function
of "((void)0)" here?

Another question, there is a paramater "x" in the former
"assert(x)", why there is not a parameter in the latter
"((void)0)"?

Cheers,
George

[Alexey B]

Sometimes you want a macro such as ASSERT to amount to nothing in specific builds, so you write code like this:#ifdef _DEBUG
#define CHECK(f, message) \
if (!(f)) \
{ \
int ID = AfxMessageBox(__FILE__ ":\n" _T(message), MB_ABORTRETRYIGNORE | MB_ICONEXCLAMATION | MB_DEFBUTTON3); \
if (ID == IDABORT) \
exit(0); \
else if (ID == IDRETRY) \
AfxDebugBreak(); \
}
#else // _DEBUG
#define CHECK(f, message) ((void)0)
#endif// _DEBUG

[cup]

What does the statement mean? What is the function
of "((void)0)" here?

It means do nothing. The alternative is

#define assert(x)

This may confuse some people but it defines assert(x) as nothing.

Another question, there is a paramater "x" in the former
"assert(x)", why there is not a parameter in the latter
"((void)0)"?

x is not used but is put in place to keep the compiler happy. assert(x) is normally used in to check whether x is non zero. If it is zero, the program will stop. This macro nullifies the effect of assert.