Re: Can we get C/C++ struct's member number, type, name, and format string?

[Mason <slash dot tmp at free dot fr>]

On 19/01/2017 03:19, hank fu wrote:

> Can we get C/C++ struct's member number, type, name, and format string?
> 
> struct c_test_struct_type {
>   int i_var;
>   unsigned int ui_var;
>   long l_var;
>   unsigned long ul_var;
> }
>
> Is there one an operator provided by GCC to determine the member
> nubmer of this struct? For example, I hope to get 4 by calling
> '__struct_member_number__(struct c_test_struct_type)' .
>
> Is there one an operator provided by GCC to get the member name of
> this struct? For example, I hope to get string 'i_var' by calling
> '__struct_member_name__(struct c_test_struct_type, 0)'.
>
> Is there one an operator provided by GCC to get the member type of
> this struct? For example, I hope to get string 'int' by calling
> '__struct_member_type__(struct c_test_struct_type, 0)'.

Challenge accepted ^_^

$ cat struct.h 
#ifdef GEN_STRUCT_DB
#define STRUCT(x)		struct { const char *type, *name; } x ## db [] =
#define MEMBER(type, name)	{ #type, #name },
#else
#define STRUCT(x)		struct x
#define MEMBER(type, name)	type name;
#endif

STRUCT(foo)
{
  MEMBER(int, i_var)
  MEMBER(unsigned int, ui_var)
  MEMBER(long, l_var)
  MEMBER(unsigned long, ul_var)
};

#undef STRUCT
#undef MEMBER

$ cat main.c 
#include "struct.h"
#define GEN_STRUCT_DB
#include "struct.h"

#define MEMBER_COUNT(x)		(sizeof x ## db / sizeof * x ## db)
#define MEMBER_TYPE(x, i)	x ## db [i].type
#define MEMBER_NAME(x, i)	x ## db [i].name

#include <stdio.h>

int main(void)
{
  unsigned i;
  struct foo bar = { 12, 34, 56, 78 };
  for (i = 0; i < MEMBER_COUNT(foo); ++i)
    printf("TYPE=%s NAME=%s\n", MEMBER_TYPE(foo, i), MEMBER_NAME(foo, i));
  return bar.i_var;
}

$ gcc -Wall -Wextra -std=c89 -pedantic -O2 main.c
$ ./a.out 
TYPE=int NAME=i_var
TYPE=unsigned int NAME=ui_var
TYPE=long NAME=l_var
TYPE=unsigned long NAME=ul_var


I know that's not what you had in mind, but I like to find ways
to (ab)use the preprocessor ;-)

Regards.