问题
When I compile the following code with the latest Visual Studio, it success to compile.
class C;
class T
{
public:
template<typename A>
void f();
private:
C* c;
};
int main()
{
T t;
t.f<int>();
}
template<typename A>
void T::f()
{
this->c->g();
}
class C
{
public:
void g() {}
};
But when I remove this->
from this->c->g()
, compilation fails with C2027: use of undefined type 'C'
.
When I make the method f
non-template, it fails to compile no matter this->
presents or not, so I think it's related to template compiling/instantiating, but I can't really figure out. I've read this answer, but isn't c
and g
unambiguous in T::f()
?
So, the question is: What's the role of this->
here?
Compiler Differences:
+-----------------------+---------------------+----------------------+--------------+
| | Template, w/ this-> | Template, w/o this-> | Non-Template |
+-----------------------+---------------------+----------------------+--------------+
| Visual Studio 16.3.10 | Success | Fail | Fail |
| x64 msvc v19.24 | Success | Success | Fail |
| x86-64 gcc 9.2 | Success w/ warning | Success w/ warning | Fail |
| x86-64 clang 9.0.0 | Fail | Fail | Fail |
+-----------------------+---------------------+----------------------+--------------+
x64 msvc v19.24
, x86-64 gcc 9.2
and x86-64 clang 9.0.0
are tested with Compiler Explorer.
回答1:
The program is ill-formed NDR due to C++17 [temp.res]/8.3:
The program is ill-formed, no diagnostic required, if:
- [...]
- a hypothetical instantiation of a template immediately following its definition would be ill-formed due to a construct that does not depend on a template parameter,
The hypothetical instantiation is ill-formed because c->g
is used when c
has pointer to incomplete type, and that is not affected by the template parameter A
.
So it is a quality of implementation issue whether an error is raised.
来源:https://stackoverflow.com/questions/60215335/template-method-accesses-forward-declared-class-fails-to-compile-only-without-th