Wierd template class

[desktop]

I am confused about the use of the template parameter "E" in the below
class. Since when is it allowed to use these parameters like "E(1)" and
what does it mean (where can I read more about this kind of use)?


template <typename E>
class Mytest {
public:
Mytest(int n) {
s = E(0);
a.resize(n,E(0));

}


void operator()() {
std::cout << "wierd operator" << std::endl;

}

int primal() {

return 34;
}

private:
E s;
std::vector<E> a;

};


The class above has a constructor "Mytest(int n)". But it also has the
freak of nature function: void operator()().

It seems that it is declaring a new operator "()" that in this case
prints "wierd operator" when called on an instance of the class, like:

Mytest<int> my(4);
my(); // executes cout in void operator block.

But if I change it to:

void operator()@ {

std::cout << "wierd operator" << std::endl;

}

I get:

main.cpp:46: error: stray ‘@’ in program

So what does the "operator" keyword really do?

 

[red floyd]

I am confused about the use of the template parameter "E" in the below
class. Since when is it allowed to use these parameters like "E(1)" and
what does it mean (where can I read more about this kind of use)?

It means construct an E with a constructor that takes an integer as a param.

If MyTest is instantiated with a class that does not have such a
constructor, the compiler should complain.

 

[Victor Bazarov]

I am confused about the use of the template parameter "E" in the below
class. Since when is it allowed to use these parameters like "E(1)"
and what does it mean (where can I read more about this kind of use)?


template <typename E>
class Mytest {
public:
Mytest(int n) {
s = E(0);
a.resize(n,E(0));

}

I would have written this constructor this way:

Mytest(int n) : s(0), a(n, E(0)) {}

void operator()() {
std::cout << "wierd operator" << std::endl;

}

int primal() {

return 34;
}

private:
E s;
std::vector<E> a;

};


The class above has a constructor "Mytest(int n)". But it also has the
freak of nature function: void operator()().

It's a "function call operator".

It seems that it is declaring a new operator "()" that in this case
prints "wierd operator" when called on an instance of the class, like:

Mytest<int> my(4);
my(); // executes cout in void operator block.

But if I change it to:

void operator()@ {

That's not C++.

std::cout << "wierd operator" << std::endl;

}

I get:

main.cpp:46: error: stray ‘@’ in program

So what does the "operator" keyword really do?

It designates the function as an overloaded operator of certain
notation.

What book are you reading that doesn't explain operator overloading?

V

 

[desktop]

I would have written this constructor this way:

Mytest(int n) : s(0), a(n, E(0)) {}


It's a "function call operator".


That's not C++.


It designates the function as an overloaded operator of certain
notation.

What book are you reading that doesn't explain operator overloading?

V

Ok the void operator()(){} now contains this code:


void operator()() {
typedef typename std::vector<E>::iterator R;
R answer = NAMESPACE::find(a.begin(), a.end(), s);
}

But R is not defined as a template parameter in the template class
definition:

template <typename E>
class Mytest {
public:
Mytest(int n) {
s = E(0);
a.resize(n,E(0));

}

void operator()() {
typedef typename std::vector<E>::iterator R;
R answer = ::find(a.begin(), a.end(), s);
}

private:
E s;
std::vector<E> a;

};

In this context I assume R is just a name for the iterator, but how can
an iterator have 2 names: "R answer = ...."?

 

[Sumit Rajan]

Ok the void operator()(){} now contains this code:


void operator()() {
typedef typename std::vector<E>::iterator R;
R answer = NAMESPACE::find(a.begin(), a.end(), s);
}

But R is not defined as a template parameter in the template class
definition:

Why should it be? Remember R is a new name for the type
std::vector<E>::iterator (the type itself is defined in std::vector). It
is *not* the name of a variable of that type.

In this context I assume R is just a name for the iterator, but how can
an iterator have 2 names: "R answer = ...."?

See above.

Regards,
Sumit.

 

[Victor Bazarov]

[..]
Ok the void operator()(){} now contains this code:


void operator()() {
typedef typename std::vector<E>::iterator R;
R answer = NAMESPACE::find(a.begin(), a.end(), s);
}

But R is not defined as a template parameter in the template class
definition:

Uh... It's defined on the line *immediately preceding* the line on
which it's used. R is a typedef-id.

template <typename E>
class Mytest {
public:
Mytest(int n) {
s = E(0);
a.resize(n,E(0));

}

void operator()() {
typedef typename std::vector<E>::iterator R;
R answer = ::find(a.begin(), a.end(), s);
}

private:
E s;
std::vector<E> a;

};

In this context I assume R is just a name for the iterator, but how
can an iterator have 2 names: "R answer = ...."?

What book are you reading that doesn't explain how 'typedef' works?

V