Question

I have looked up how to pass an enumeration in a function, but that method doesn’t work when both the function and enumeration are declared in a structure. Here is my code :

test_setup.h:

     1 #ifndef TEST_SETUP_H_
     2 #define TEST_SETUP_H_
     3 #include <stdio.h>
     4 #include <stdlib.h>
     5 
     6 
     7 typedef struct _test_setup {
     8 
     9   int *partner;            
    10   int *send_first;         
    11   double *results;         
    12 
    13   enum { CHIP_WIDE, NODE_WIDE, SYSTEM_WIDE } SomeMatches;
    14   void match_partners(SomeMatches match);
    15 
    16 } test_setup;              
    17 
    18 #endif

test_setup.c :

     1 #include "../includes/test_setup.h"
     2 
     3 void match_partners(SomeMatches match) { 
     4   if (match == CHIP_WIDE) {
     5 
     6   }
     7   else if (match == NODE_WIDE) {
     8 
     9   }
    10   else if (match == SYSTEM_WIDE) {
    11 
    12   }
    13   else {
    14 
    15   }
    16 }`

Error:

    In file included from src/test_setup.c:1:
    src/../includes/test_setup.h:14: error: expected ‘)’ before ‘match’
    src/../includes/test_setup.h:16: warning: no semicolon at end of struct or union
    src/test_setup.c:3: error: expected ‘)’ before ‘match’
    make: *** [setup.o] Error 1

I have tried every combination of declaring an enumeration and using it in the function parameters, but nothing has worked. Any ideas would be much appreciated. I am compiling with mpicc (because the rest of the program uses MPI functions) but I have tried with GNU GCC and I get the same warnings/errors.

Answer 1

For C

If you truly want C, then you simply can’t do any of this.

• You can’t define a member function in a struct
• You can’t define a named enumeration nested in a struct

For C++

Use the scope-resolution operator ::

#include<iostream>

struct Test
{
  enum SomeEnum { TEST1, TEST2, TEST3 };
  void SomeFunction(SomeEnum e);
};

void Test::SomeFunction(Test::SomeEnum e)
{
  std::cout << e << "\n";
}

int main(int argc, char* argv[])
{
  Test t;
  t.SomeFunction(Test::TEST1);
  return 0;
}