1

Consider the following template:

template <typename T>
void foo() {
  static_assert(sizeof(T) == 0, "Now what?");
}

The standard (§7.4) says:

[If the condition to static_assert is false] the program is ill-formed, and the resulting diagnostic message (1.4) shall include the text of the string-literal, […]

(Reference taken from https://stackoverflow.com/a/11048906/560450)

In practice, the static_assert will not fail until we instantiate the function template foo, because the use othe the template parameter forces the compiler to evalute the condition only during two-phase lookup. Therefore, the code snippet above is not deemed ill-formed unless the function template is instantiated.

On the other hand, sizeof(T) > 0 in C++ by definition. Therefore, the value of the condition is in fact independent of any template parameter! Is a "malicious" compiler allowed to take advantage of this fact, and reject the program as ill-formed, regardless whether foo is actually instantiated or not?

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Kristóf Marussy
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1 Answers1

7

Yes, the compiler is allowed, but not required, to reject this.

§14.6 [temp.res]/p8:

If no valid specialization can be generated for a template, and that template is not instantiated, the template is ill-formed, no diagnostic required.

An easy workaround:

template <typename T>
struct foo_protector : std::false_type {};

template <typename T>
void foo() {
  static_assert(foo_protector<T>::value, "Now what?");
}

The compiler is not allowed to reject foo()'s definition because there might be a specialization of foo_protector, not yet seen, such that its value member is true. As long as you don't actually write such a specialization, the static_assert will fire as expected.

T.C.
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  • Nice find, thanks! Now we only need to wait for C++ compilers equipped with automated theorem provers (and while we are at it, let's throw in a halting oracle for good measure) for this to happen in non-trivial circumstances. :-) – Kristóf Marussy Dec 07 '14 at 14:55