C / C++ / MFC

Maxwell Chen wrote:

1. You saw the keyword template.

2. A struct is a class with members being public by default.

George_George wrote:

I can not believe when Bjarne said "Even if a template is first used within a class", he means even if a template keyword? is first used within a class? It should be some specific template, right?

Not can an unqualified name used in a template
 ever be bound to a local name. 

Finally, even if a template is first used within a class,
 unqualified names used in the template will
 not be bound to members of that class.

template<classT>
void sort(vector<T>& v)
{
  sort(v.begin(), v.end()); // use standard library sort()
}
class Container{
  vector<int> v; // elements
  //...
public:
  void sort() // sortelements
  {
    sort(v); // invokes sort(vector<int>& ) rather than Container::sort()
  }
  //...
};

Maxwell Chen wrote:

Not can an unqualified name used in a template ever be bound to a local name.

template <typename T> struct Base {
    int f();
};

template <typename T> struct Derived : Base<T> {
    int g() { return f(); };
};

Maxwell Chen wrote:

Finally, even if a template is first used within a class, unqualified names used in the template will not be bound to members of that class.

BadKarma wrote:

Even when there are global variables or function is correctly understand what the programmer wants and does a check the base class first.

BadKarma wrote:

1. Can one write code which breaks this rule? Meaning, can on write code that explicit calls the global function instead of the base function?

#include <iostream>

using namespace std;

static int i = 100;

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.i << endl; // output 200
	cout << d.get_i() << endl; // output 100
	return 0;
}

George_George wrote:

Sure, we can. Here is the code to prove.


#include <iostream>

using namespace std;

static int i = 100;

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.i << endl; // output 200
cout << d.get_i() << endl; // output 100
return 0;
}

int get_i() { return ::i; }

George_George wrote:

What do you mean this before?

"Even when there are global variables or function is correctly understand what the programmer wants and does a check the base class first."

BadKarma wrote:

the VC compilter correctly choses the i defined in the base class.

namespace N {
   void f(int) { printf("f(int)\n");}
}
 
template <class T> void g(T) {
   N::f('a');   // calls f(char) should call f(int)
}
 
namespace N {
   void f(char) { printf_s("f(char)\n");}
}
 
int main() {
   g('c');
}

template <class T> void g(T t) {
   N::f(t);
}

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>        ]

#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;
}

nitin3 wrote:

How to seperate main windows from them extactly as Windows TaskManager does ?

int get_i() { return i; }

Maxwell Chen wrote:

because my domain server blocks the copy-and-paste function