Compile-time assert when not all enum values are handled in a switch statement in C++

Question

I'd like to get a compiler warning or error when not all possible enum values are handled in a switch statement. Of course I can add a default case with an assert and (eventually) get an error at runtime. But I'd like to get an error at compile-time.

I'm not sure if this is possible at all with C++, but maybe someone knows a trick...

Edit: Using -Wswitch seems to be the solution for GCC. Is there something similar for VS2010? (I'm not using GCC).

Edit2: Ok, I found the solution for VC++ (VS2010):

Enabling warning C4062 produces a warning when a value is missing und no default case is provided.

Enabling warning C4061 produces a warning when a value is missing, even if a default case is provided.

Answer 1

You haven't mentioned which compiler you're using. If you're using GCC, you can get that for free simply by enabling -Wswitch (which is automatically enabled by -Wall).

Answer 2

AFAIK there's no conventional way to achieve what you want with MSVC. There're tricks to do similar things, but they involve either a sophisticated template voodoo, or really fierce macros riddles.

For example, instead of defining your enum in a conventional way do the following:

#define MyEnumEntries(m) \
    m(A, 1) \
    m(B, 2) \
    m(C, 3) \

enum Abc {

    // the following will expand into your enum values definitions
#   define M_Decl(name, value) name = value,
    MyEnumEntries(M_Decl)
};

Now, your switch can be rewritten into this:

Abc a = A;

switch( a )
{
#define M_Goto(name, value) \
case value:
    goto label_##name;

MyEnumEntries(M_Goto)


case label_A:
    // TODO
    break;
case label_B:
    // TODO
    break;
}

The above will not compile if you won't add the switch entry label_... for all the enum values.