Why doesn't gcc allow a const int as a case expression?

I was looking at this SO question and got to thinking about const ints versus #defines and realized I don't actually understand why the compiler can't deal with this. Could someone shed some light as to why the following code

const int FOO = 10;

int main(int argc, char** argv)
{
    switch(argc)
    {
        case FOO: { printf("foo\n"); }
        default:  { printf("default\n"); }
    }
}

results in

error: case label does not reduce to an integer constant

I read the ISO-C99 spec which states in 6.8.4.2.3 that

The expression of each case label shall be an integer constant expression and no two of the case constant expressions in the same switch statement shall have the same value after conversion.

I understand why the case expression must be constant, but not why only a literal makes the compiler (gcc 4.2.1) happy.