static assert for const variables?

Question

Static asserts are very convenient for checking things in compile time. A simple static assert idiom looks like this:

template<bool> struct StaticAssert;
template<> struct StaticAssert<true> {};

#define STATIC_ASSERT(condition) do { StaticAssert<(condition)>(); } while(0)

This is good for stuff like

STATIC_ASSERT(sizeof(float) == 4)

and:

#define THIS_LIMIT (1000)
...
STATIC_ASSERT(THIS_LIMIT > OTHER_LIMIT);

But using #define is not the "C++" way of defining constants. C++ would have you use an anonymous namespace:

namespace {
    const int THIS_LIMIT = 1000;
}

or even:

static const int THIS_LIMIT = 1000;

The trouble with this is that with a const int you can't use STATIC_ASSERT() and you must resort to a run-time check which is silly.

Is there a way to properly solve this in current C++?
I think I've read C++0x has some facility to do this...

---

EDIT

Ok so this

static const int THIS_LIMIT = 1000;
...
STATIC_ASSERT(THIS_LIMIT > 0);

compiles fine
But this:

static const float THIS_LIMIT = 1000.0f;
...
STATIC_ASSERT(THIS_LIMIT > 0.0f);

does not.
(in Visual Studio 2008)

How come?

Answer 1

Why, you can still static assert with const int:

#define static_assert(e) extern char (*ct_assert(void)) [sizeof(char[1 - 2*!(e)])]
static_assert( THIS_LIMIT > OTHER_LIMIT )

Also, use boost!

BOOST_STATIC_ASSERT( THIS_LIMIT > OTHER_LIMIT )

... you'll get a lot nicer error messages...

Answer 2

static_assert is a compiler feature in C++0x so as long as you've got a relatively up-to-date compiler you can use that. Watch out for doing #define static_assert(x) ..., because it's a real keyword in C++0x so you'd be permanently hiding the compiler feature. Also, C++0x static_assert takes two parameters (eg. static_assert(sizeof(int) == 4, "Expecting int to be 4 bytes")), so you could cause yourself problems trying to switch in future if you use that #define.