What's the meaning of the macro 'OF'

[StdNewer]

#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args
# else
# define OF(args) ()
# endif
#endif

void func OF( args );

I saw some C programs written like this. Why?

[Ian Collins]

StdNewer wrote:

#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args

Are you sure that wasn't define OF(args) (args)?

# else
# define OF(args) ()
# endif
#endif

void func OF( args );

I saw some C programs written like this. Why?

Very old (before 1989) compilers didn't have function prototypes.
Expand the macros and see what you get with and without STDC defined.

--
Ian Collins.

[Ben Pfaff]

Ian Collins <ia******@hotma il.comwrites:

StdNewer wrote:

>#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args

Are you sure that wasn't define OF(args) (args)?

Probably not: then it would only work with 0- or 1-parameter
functions, because a 1-argument macro cannot be invoked with
multiple comma-separated arguments.
--
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={0x67dffdff,0x 9aa9aa6a,0xa77f fda9,0x7da6aa6a ,0xa67f6aaa,0xa a9aa9f6,0x11f6} ,*p
=b,i=24;for(;p+ =!*p;*p/=4)switch(0[p]&3)case 0:{return 0;for(p--;i--;i--)case+
2:{i++;if(i)bre ak;else default:continu e;if(0)case 1:putchar(a[i&15]);break;}}}

[StdNewer]

On 2ÔÂ2ÈÕ, ÏÂÎç2ʱ42·Ö, Ian Collins <[e-mail] mwrote:

StdNewer wrote:

#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args

Are you sure that wasn't define OF(args) (args)?

I see,thanks very much.
I wrote wrong in "void func OF(args)"; it should be: void func
OF((args));
But if "#define OF(args) ()",this function is vod fun(()).
two parentheses,rig ht?
why didn't he define "OF" as you do,and declare function as
func OF(),not func OF(())?

[Ian Collins]

StdNewer wrote:

On 2ÔÂ2ÈÕ, ÏÂÎç2ʱ42·Ö, Ian Collins <[e-mail] mwrote:

>StdNewer wrote:

>>#ifndef OF /* function prototypes */
# ifdef STDC
# define OF(args) args

Are you sure that wasn't define OF(args) (args)?

I see,thanks very much.
I wrote wrong in "void func OF(args)"; it should be: void func
OF((args));
But if "#define OF(args) ()",this function is vod fun(()).
two parentheses,rig ht?
why didn't he define "OF" as you do,and declare function as
func OF(),not func OF(())?

See Ben's reply.

--
Ian Collins.