Compiler Robustness ... Naivete - c++

Question

I use Visual Studio Professional 2012. I pre-compiled a class (header and source) successfully. Days later, when compiling another class (for the moment header only) that is utilizing the previous one, the compiler caught a missing reference if(this != &rhs) and semicolon rhs.root = nullptr;.
Perhaps it is my naivete and lack of knowledge about how compilers work but I thought a compiler was robust to catch errors such as these. It appeared to me that only when a specific block of code was required did the compiler feel the need to check it.
I have read about just-in-time compilation and learned how assembly compilers perform a two-pass compilation with symbols first and then syntax. I have not taken a course in compiler construction at my university and I know such courses give great insight into parsers, etc.
The code section where it failed to catch the error is this move assignment operator:
Tree &operator=(Tree &&rhs)
{
if(this != rhs) <--------- no reference to the rhs
{
root = std::move(rhs.root);
rhs.root = nullptr <----------- no semicoln
}
return *this;
}
The errors were generated while compiling boost variant, as well as my visitor class member:
bool operator() (Tree<std::string>& tree) const {
return tree.load(tree);
}
as well as a host of other errors related to boost serialization. The fix was to, of course, correct the missing reference and semicolon but I want to understand why this was caught apparently only when the compiler needed to touch this code?

Answer 1

Is it a template class?
Because semantic analysis of templates makes only sense when they are instantiated. That is, if it is a template, the compiler should generate an error at the missing semicolon (syntactic error), but not at the == operator.
The following code compiles with g++:
template<typename T>
struct A {
void q(A &a) {
if (this == a) {}
}
};
int main(int argc, char **argv) {
A<int> x;
//x.q(x);
}
But doesn't compile when
x.q(x);
is uncommented.