Passing Enumerations in a Function that's in a Struct (in C) - c

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;
}

Answer 2

If you want to do this in C, here are the workarounds:
// define the enumeration at file scope:
typedef enum {CHIP_WIDE, NODE_WIDE, SYSTEM_WIDE} SomeMatches;
// declare a *pointer* to a function in the struct:
typedef struct {
int *partner;
int *send_first;
double *results;
void (*match_partners)(SomeMatches match);
} test_setup;
Define your function as normal:
void match_partners(SomeMatches match)
{
if (match == CHIP_WIDE) {}
else if (match == NODE_WIDE) {}
else if (match == SYSTEM_WIDE) {}
}
Then when you create an instance of the struct, assign the function pointer:
test_setup t;
t.match_partners = match_partners;
You don't need to explicitly dereference the function pointer to call it, so you can execute your function as
t.match_partners(CHIP_WIDE);
although if you want to dereference it explicitly, use
(*t.match_partners)(CHIP_WIDE);
Note that C doesn't have any equivalent to the this pointer; if match_partners depends on information contained in the struct instance, you'll have to explicitly pass that instance as a separate argument:
void match_parthers(SomeMatches matches, test_setup *instance)
{
}
...
typedef struct {
...
void (*match_partners)(SomeMatches matches, test_setup *instance);
...
} test_setup;
...
test_setup t;
t.match_partners = match_partners;
t.match_partners(CHIP_WIDE, &t);
Note that for the struct definition to be legal, we have to pass the instance as a pointer, since the test_setup type isn't complete at that point.
EDIT
That last sentence isn't terribly clear; let me try again. A struct definition cannot refer to an instance of itself, because the struct type isn't complete until the closing }. IOW, the following is not legal:
struct foo
{
...
struct foo bar;
...
};
However, a struct can refer to a pointer to another instance of the same type, so the following is legal:
struct foo
{
...
struct foo *bar;
...
};
The same logic applies to the function pointer declaration; the instance parameter needs to be declared as test_setup *, since the test_setup type definition isn't complete at that point. As Merlyn points out in the comments, you probably want the instance to be mutable anyway, but the language requires you to use a pointer in that circumstance.

Answer 3

Your enum does not have a proper name (it doesn't have a tag) and cannot be referenced from distinct points in the code. Try this:
enum SomeMatches { CHIP_WIDE, NODE_WIDE, SYSTEM_WIDE };
struct _test_setup {
/* ... */
enum SomeMatches SomeMatches;
void (*match_partners)(enum SomeMatches match);
/* ... */
};
Sprinkle typedefs at will, or leave things bare ...

Answer 4

First of all #define TEST_SETUP_H_ goes before #ifndef TEST_SETUP_H_. Otherwise your code will be completely commented out to the compiler.
You don't name a typedef struct like that it goes as such:
typedef struct struct_name{
/contents/
};
void match_partners(SomeMatches match) can be void match_partners(void) and use the variable inside the function because with that header file you've made it global and don't need to pass it.