What is the role of this macro in function declaration?

Question

I have downloaded some library and it declares the functions the following way:

#if !defined(__ANSI_PROTO)
#if defined(_WIN32) || defined(__STDC__) || defined(__cplusplus)
#  define __ANSI_PROTO(x)       x
#else
#  define __ANSI_PROTO(x)       ()
#endif
#endif

int httpdAddVariable __ANSI_PROTO((httpd*,char*, char*));

What is the role of __ANSI_PROTO here? Why it is preferred to declaring simply as

int httpdAddVariable (httpd*,char*, char*);

it a macros for declaring functions and arguments with or without * ANSI C prototypes — Seek Addo, Jan 03 '17 at 15:59
Perhaps to cope with early C compilers which lacks argument signature? — chux - Reinstate Monica, Jan 03 '17 at 15:59

Petr Skocik · Accepted Answer · 2017-01-03T16:09:06.930

7

Pre-ANSI C didn't support this:

int httpdAddVariable (httpd*,char*, char*);

It only supported this:

int httpdAddVariable (); /* = arguments unspecified*/

So that's what the macro does. It pastes the argument types into the declaration if it detects prototype support; otherwise, it just pastes ().

edited Jan 03 '17 at 16:09

answered Jan 03 '17 at 16:02

Petr Skocik

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So I can safely remove all appearances of this macro from the code? – Marin Rantic Jan 03 '17 at 16:05
4

Yes. (Unless you plan on supporting ancient compilers). – Petr Skocik Jan 03 '17 at 16:06

What is the role of this macro in function declaration?

1 Answers1