未实例化未定义的 class 模板以检查友元函数

Question

The following program compiles (see on godbolt ), but it would not compile if we uncommented the definition of Buffer .

template <int size>
struct Buffer /*{ char buf[size]; }*/;

template <class T>
struct Wrapper { void operator+() {} };

Wrapper<Buffer<-5>> a;

void f() { +a; }

The reason, the uncommented version does not compile: +a triggers ADL, and to collect all candidates for operator+ , all associated classes must be checked for friend functions. Buffer<-5> is an associated class, so it must be instantiated. Instantiation fails, hence the compilation error. See this question .

I wonder if Buffer<-5> must be instantiated, why don't we have a compilation error, if Buffer is not defined?

Answer 1

You can (implicitly) instantiate a class template from only a declaration; you get an incomplete class type, just like from struct A; ( [temp.inst]/2 ). It's not an error, of course, to do ADL with an incomplete associated class; the class in question is simply not searched for friend declarations.