Re: anti-standard code can compile -- about template class
George wrote:
Thanks Bo Persson,
Can you explain why if we comment the line cout << d.get_i() <<
endl; in main, there will not be compile error, but if we invoke
cout << d.get_i() << endl;, there will be compile error?
From your points (I accept currently I agree), I can not explain.
:-)
If you actually call get_i(), the compiler is forced to look for an
'i' and discovers that it cannot find one.
If it worked properly (two phase lookup), it should have discovered
that already when compiling the definition of get_i().
Bo Persson
Any comments or ideas?
Compile errors,
1>------ Build started: Project: test_template4, Configuration:
Debug Win32 ------
1>Compiling...
1>main.cpp
1>d:\visual studio
2008\projects\test_template4\test_template4\main.cpp(20)
error C2065: 'i' : undeclared identifier
1> d:\visual studio
2008\projects\test_template4\test_template4\main.cpp(20) : while
compiling class template member function 'int
Derived<T>::get_i(void)' 1> with
1> [
1> T=int
1> ]
1> d:\visual studio
2008\projects\test_template4\test_template4\main.cpp(26) : see
reference to class template instantiation 'Derived<T>' being
compiled 1> with
1> [
1> T=int
1> ]
[Code]
#include <iostream>
using namespace std;
template <typename T> struct Base {
public:
Base (int _i): i (_i)
{
}
int i;
};
template <typename T> struct Derived : public Base<T> {
public:
Derived (int _i) : Base<T> (_i)
{
}
int get_i() { return i; }
};
int main()
{
Derived<int> d (200);
cout << d.get_i() << endl; // output 200
return 0;
}
[/Code]
regards,
George