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__) 4 2037
On Sat, 24 Jul 2004 22:39:45 +0200, Leo <le*******@emai l.com> wrote: 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
John Harrison wrote: On Sat, 24 Jul 2004 22:39:45 +0200, Leo <le*******@emai l.com> wrote:
.... 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
John Harrison wrote: On Sat, 24 Jul 2004 22:39:45 +0200, Leo <le*******@emai l.com> wrote:
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
Both versions won't compile :D
On Sun, 25 Jul 2004 00:38:29 +0200, Leo <le*******@emai l.com> wrote: John Harrison wrote: On Sat, 24 Jul 2004 22:39:45 +0200, Leo <le*******@emai l.com> wrote:
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
Both versions won't compile :D
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 This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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