Question

struct info _info;

#define INIT(a, b, c, d)                   \
    struct info _info = {a, b, c, d}

This worked in C, but with the g++ I get:

error: redefinition of ‘info _info’

INIT isn’t always called, sometimes _info gets initialised some other way so that’s why both ways have to stick around.

Context:

I’m using INIT in a file that is getting compiled with g++, but I also use it in files compiled by gcc. So the problem is: I need this headerfile code to work in both languages, regardless of whether I use the header in a c library or in a c++ library.

Kerrek pointed out I could use #ifdef, so I did this:

#ifndef __cplusplus

struct info _info;
#define INFO(a, b, c, d)                   \
    struct info _info = {a, b, c, d}

#else

struct info _info;
#define INFO(a, b, c, d)                   \
            _info = {a, b, c, d}
#endif

But it still won’t work, I’m getting a error: ‘_info’ does not name a type
at the line I’m using the macro in my cpp project:
INFO("Information", 2 ,"again", 4)

Answer 1

In C++, you don’t say struct in the variable declaration. The following should work:

struct info { int x, y, z, w; };  // type definition (elsewhere in your code)

#define INIT(a, b, c, d) info _info = {a, b, c, d}

INIT(1,2,3,4);

Because the variable name is fixed, this macro can only be used once inside any given scope, which is not obvious. For more flexibility, I’d add the variable name to the macro:

#define DECLARE_AND_INIT(var, a, b, c, d) info var = {a, b, c, d}
DECLARE_AND_INIT(my_info, 4, 5, 6, 7);