Using data types in an enum declaration

Question

Can I do something like this? I would like to use data types instead of constants in my enum type_t.

typedef struct {
    char id;
    long data;
} type1_t;

typedef struct {
    char id;
    long data;
    float moredata;
} type2_t;

typedef enum {
    type1_t, type2_t
} type_t;

typedef struct {
    type_t type;
    char* something;
} midas;

midas obj1;
obj1.type = type1_t;
obj1.type.id = 0;
obj1.type.data = 123;

midas obj2;
obj2.type = type2_t;
obj2.type.id = 3;
obj2.type.data = 456;
obj2.type.moredata = 3.14;

In the example the type variable of the midas struct should then refer to type1_t or type2_t. So if I set the type to type2_t, the size of it should be bigger than when I set type1_t.

cdhowie · Accepted Answer · 2010-11-10 20:58:58Z

2

No, you can't do that. If you were using C++, you could use templates to achieve this. But there is no "type enum" mechanism in C.

You could consider a union:

typedef enum {
    type1, type2
} type_t;

typedef struct {
    type_t typecode;
    union {
        type1_t type1value;
        type2_t type2value;
    };
    char* something;
} midas;

This will cause midas.type1value and midas.type2value to occupy the same memory space. The amount of memory taken for the union will be equal to the amount of memory required to store the largest data type it contains.

You would then have to look at midas.typecode and consider which union member to use. If you use the wrong one you will wind up with invalid data and this may lead to program crashes, so be careful.

edited Nov 10, 2010 at 20:58

answered Nov 10, 2010 at 20:51

cdhowie

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2 Comments

Midas Over a year ago

Thanks, that was a quick response! Isn't there a way to allocate memory for the union, or isn't that needed?

cdhowie Over a year ago

The memory is allocated as part of the midas struct itself. The regions of memory used will just overlap, so you should only ever use one or the other for any given instance of the struct. You can demonstrate this by asserting: midas value; if (&(value.type1value) == &(value.type2value)) printf("The members are in a union.\n"); You will see that pointers to each member of the union are equal in value.

Adam Rosenfield · Accepted Answer · 2010-11-10 20:51:27Z

2

No. The usual way to do something like this in C is with a union:

typedef enum { TYPE1, TYPE2 } type_t;

typedef struct {
    type_t type;
    union
    {
        type1_1 t1;
        type2_t t2;
    } u;
} midas;

midas obj1;
obj1.type = TYPE1;
obj1.u.t1.id = 0;
obj1.u.t1.data = 123;

midas obj2;
obj2.type = TYPE2;
obj2.u.t2.id = 3;
obj2.u.t2.data = 456;
obj2.u.t2.moredata = 31.14;

answered Nov 10, 2010 at 20:51

Adam Rosenfield

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2 Comments

Ben Jackson Over a year ago

The union can be anonymous to avoid the u. everywhere. Before that was widely supported it was common to have blocks of #define t1_id u.t1.id

Adam Rosenfield Over a year ago

@Ben Jackson: That's true, but that's not standard C, it's an extension to C provided by compilers. If you want completely portable code, you should not use anonymous unions.

Ben Jackson · Accepted Answer · 2010-11-10 20:50:17Z

1

I think you are looking for union

answered Nov 10, 2010 at 20:50

Ben Jackson

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Comments

Oliver Charlesworth · Accepted Answer · 2010-11-10 20:52:31Z

1

No, you can't do that. enums are more or less equivalent to ints.

You can do this sort of thing was a union, e.g.:

typedef struct
{
    int   a;
    float b;
} type1_t;

typedef struct
{
    char  c;
    long  d;
} type2_t;

typedef union
{
    type1_t  t1;
    type2_t  t2;
} myUnion_t;


...

myUnion_t u;
u.t1.a = 3;
u.t1.b = 2.7f;

answered Nov 10, 2010 at 20:52

Oliver Charlesworth

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Comments

frast · Accepted Answer · 2010-11-10 20:50:33Z

0

That does not work. The closest thing would be to use a void* to your structure.

answered Nov 10, 2010 at 20:50

frast

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