assert-macro: brackets

[Leo]

Hi @ all.

I looked up the implementation of the assert macro of my compiler
(MinGW), because I wanna write my own assert.

I found this:

#define assert(x) ((void)0)
#define assert(e) ((e) ? (void)0 : _assert(#e, __FILE__, __LINE__))

My question is: Do the outer brackets have any reason? If yes, which?

Is this not equivalent to the above code?:

#define assert(x) (void)0
#define assert(e) (e) ? (void)0 : _assert(#e, __FILE__, __LINE__)

 

[John Harrison]

Hi @ all.

I looked up the implementation of the assert macro of my compiler
(MinGW), because I wanna write my own assert.

I found this:

#define assert(x) ((void)0)
#define assert(e) ((e) ? (void)0 : _assert(#e, __FILE__, __LINE__))

My question is: Do the outer brackets have any reason? If yes, which?

Is this not equivalent to the above code?:

#define assert(x) (void)0
#define assert(e) (e) ? (void)0 : _assert(#e, __FILE__, __LINE__)

No, try the following with your two versions of assert

if (0 && assert(0))
{
}

With the first definition this will not call _assert (which is correct),
with the second definition it will.

john

 

[Gianni Mariani]

....

No, try the following with your two versions of assert

if (0 && assert(0))

I think you meant

if (0 && assert(1))

{
}

With the first definition this will not call _assert (which is
correct), with the second definition it will.

G

 

[Leo]

No, try the following with your two versions of assert

if (0 && assert(0))
{
}

With the first definition this will not call _assert (which is
correct), with the second definition it will.

john

Both versions won't compile

 

[John Harrison]

Both versions won't compile

Yes, I realised that after I posted.

How about this?

1 && assert(0);

That fails to compile with version 1, but compiles with version 2.

I'm sure a better example could be devised, but the point is that the
extra bracket in the first version avoids any surprises.

john