请帮助我理解这种语法（在C ++中实现静态断言）

Question

This syntax was used as a part of an answer to this question :

template <bool>
struct static_assert;

template <>
struct static_assert<true> {}; // only true is defined

#define STATIC_ASSERT(x) static_assert<(x)>()

I do not understand that syntax. How does it work?

Suppose I do

STATIC_ASSERT(true);

it gets converted to

static_assert<true>();

Now what?

Answer 1

STATIC_ASSERT(true);

indeed means

static_assert<true>();

which evaluates to nothing. static_assert<true> is just an empty structure without any members. static_assert<true>() creates an object of that structure and does not store it anywhere.

This simply compiles and does nothing.

On the other hand

STATIC_ASSERT(false);

means

static_assert<false>();

which results in compilation error. static_assert has no specialization for false . So a general form is used. But the general form is given as follows:

template <bool>
struct static_assert;

which is just a declaration of a structure and not its definition. So static_assert<false>() causes compilation error as it tries to make an object of a structure which is not defined.

Answer 2

static_assert<true>(); makes that

template <>
struct static_assert<true> {}

templated struct specialization temporary object creation being done - a call to constructor and later to a destructor that both will be hopefully eliminated by the optimizer since they do nothing. Since there's only a specialization for true and no generic version of the template struct all constructs that evaluate to static_assert<false>(); will just not compile.

Answer 3

In the expression

static_assert<true>();

since static_assert<true> is a type, it will call the constructor of static_assert<true> . As static_assert<true> is specialized to an empty struct, nothing will be affected.

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However, in

static_assert<false>();

as there is no specialization for static_assert<false> , the generic definition

template <bool>
struct static_assert;

will be used. But here, the type static_assert<B> is incomplete . So calling constructor of static_assert<B> will result in compilation error.

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Therefore, this is called "static assert" as the statement will abort the compilation if the expression evaluates to false , similar to the normal assert() function that will kill the program in runtime.

Answer 4

Well, I guess it is about template specialization. STATIC_ASSERT(true) will compile successfully, because there is a definition (not just a declaration) of "static_assert< true >".

STATIC_ASSERT(false) will be rejected by the compiler, because there is only a declaration for "static_assert< false >" and no definition.

Update: for visual studio, STATIC_ASSERT(true) is ok, but STATIC_ASSERT(false) triggers the error: "error C2514: 'static_assert<__formal>' : class has no constructors [ with __formal = false ]"