A question about #define

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

Sometimes you want a macro such as ASSERT to amount to nothing in specific builds, so you write code like this:

Code:

---

#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

---

Quote:

---

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.

Quote:

---

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.