问题
I'm reviewing a codebase where assert macro is expanded like this in non-debug configurations:
#define assert( what ) while( 0 )( ( void )1 )
which I don't quite get. Obviously the goal is to have a no-op. Then why not expand into an empty string?
#define assert( what )
What's the reason for this no-op loop?
回答1:
Most likely to avoid compiler warnings. Check whether this code provokes a warning about an empty statement:
if (foo);
If it does, then do you want the same warning in release mode for the following code?
if (foo) assert(what);
C99 (which is relevant to C++11) also says that assert expands "to a void expression". IIRC, whitespace alone isn't an expression, even though whitespace followed by a semi-colon is an expression-statement. Good old BNF grammar.
By the way, this definition of assert is not standard-conforming. C89 and C99 both say that when NDEBUG is defined, then assert is defined as:
#define assert(ignore) ((void)0)
I'm not sure whether the authors consider it an important requirement, but a program could for example stringify the macro expansion and expect a particular result.
回答2:
To gobble up the semicolon, most likely. Whether it's needed or not — I don't think it'll make much difference, but at the same time, it doesn't hurt anything, either.
来源：https://stackoverflow.com/questions/7966967/whats-the-reason-for-this-no-op-while-loop-used-for-assert-macro