如果声明了类型，则为 static_assert

Question

In c++11 is it possible to cause a static_assert to fail if a type is declared. And by declared I mean whether forward declared or fully defined.

The purpose is to fail a compile with an instructive message if something has or has not been declared already.

Excepting a clever trick, I fully expect that some intermediate type_traits-like template to be involved which isn't already part of the standard.

Answer 1

If a type does not exist at a given point in the code, then you can't refer to it there. In order to refer to it you'd have to declare it. But then you can't know which declaration any use of it refers to. The only possibility to effectively assert that it doesn't exist, is then to declare or define it in a way that would clash with an earlier declaration, eg like this:

struct Type;    // Earlier declaration.

// Whatever, then:
using Type = struct Unique_temporary_name*;    // "Type" must not exist.

Answer 2

You can statically assert that its size is what you expect.

typedef struct _ListNode {
    struct _ListNode* next;
    int value;
} ListNode;

static_assert(sizeof(ListNode) >= sizeof(void*) + sizeof(int));

If the type exists, this will work the same as any other static_assert(…) , if not better since this also sanity checks the size of the type. (If you didn't want that, you could simply static_assert(sizeof(ListNode) >= 1) . If the type does not exist, you will get a regular error, not an assertion failure, but I don't think that can be avoided.

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_{Nit: The >= is needed due to padding bytes added by the compiler, unless you were to include #pragma pack(1).}