How do you do assignment operator within a template ?

[imutate]

How do you do assignment operator within a template ?

#include <vector>

template < typename T >
class Vec : public std::vector< T {
public:
Vec() { }
Vec( int s ) : std::vector<T>(s) { }
T& operator[](int i) { return this -at(i); }
const T& operator[](int i) const { return this -at(i); }
T& operator=(const T& rhs) { return this -// and then what, if
that is right atall ? }
};

[Boris]

im*****@hotmail.co.uk wrote:

How do you do assignment operator within a template ?

#include <vector>

template < typename T >
class Vec : public std::vector< T {
public:
Vec() { }
Vec( int s ) : std::vector<T>(s) { }
T& operator[](int i) { return this -at(i); }
const T& operator[](int i) const { return this -at(i); }
T& operator=(const T& rhs) { return this -// and then what, if
that is right atall ? }
};

What shall your assignment operator do? If you have a vector with 10 strings
and you assign a new string to your vector class shall it replace all the 10
strings?

Boris

[Victor Bazarov]

im*****@hotmail.co.uk wrote:

How do you do assignment operator within a template ?

#include <vector>

template < typename T >
class Vec : public std::vector< T {
public:
Vec() { }
Vec( int s ) : std::vector<T>(s) { }
T& operator[](int i) { return this -at(i); }
const T& operator[](int i) const { return this -at(i); }
T& operator=(const T& rhs) { return this -// and then what, if
that is right atall ? }
};

Not sure what it is you're asking. What do you need your operator
to do? Why are you assigning from an object to T to a vector? Do you
intend to assign all of the elements to that value? Then you need
a loop.

V
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[Gianni Mariani]

im*****@hotmail.co.uk wrote:

How do you do assignment operator within a template ?

#include <vector>

template < typename T >
class Vec : public std::vector< T {
public:
Vec() { }
Vec( int s ) : std::vector<T>(s) { }
T& operator[](int i) { return this -at(i); }
const T& operator[](int i) const { return this -at(i); }
T& operator=(const T& rhs) { return this -// and then what, if
that is right atall ? }

somthing like this should work.
T& operator=(const T& rhs) {
{
// this->thing = rhs.thing;
... etc
return *this;
}

};

[Greg]

oh sorry I did not think of that case, no I wanted to see when elements
are being assigned to

T& operator=[](const &i, const T& rhs) {
{
this->thing[i] = rhs.thing;
std::cout << "its here << " i << " " << rhs.thing;
return *this;
}

[Victor Bazarov]

Greg wrote:

oh sorry I did not think of that case, no I wanted to see when
elements are being assigned to

T& operator=[](const &i, const T& rhs) {
{
this->thing[i] = rhs.thing;
std::cout << "its here << " i << " " << rhs.thing;
return *this;
}

No, you can't do that. Indexing operator has only one argument.

If you want to see when your elements are assigned to, do not provide
the indexing operator but instead have a 'set' type function:

void set(int i, const T& val);

V
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[Greg]

No, you can't do that. Indexing operator has only one argument.

>
If you want to see when your elements are assigned to, do not provide
the indexing operator but instead have a 'set' type function:

void set(int i, const T& val);

Will the assignment call this "set" e.g.

std::vector<intx;
x[0] = blah;

or would i have to replace "=" by set function ?
x.set(0, blah);

[Victor Bazarov]

Greg wrote:

>No, you can't do that. Indexing operator has only one argument.

If you want to see when your elements are assigned to, do not provide
the indexing operator but instead have a 'set' type function:

void set(int i, const T& val);

Will the assignment call this "set" e.g.

std::vector<intx;
x[0] = blah;

or would i have to replace "=" by set function ?
x.set(0, blah);

The latter.

If you need to track assignment, you would have to implement a proxy
class. See my recent post to comp.lang.c++.moderated for an example.

V
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[Greg]

The latter.

>
If you need to track assignment, you would have to implement a proxy
class. See my recent post to comp.lang.c++.moderated for an example.

V
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OK thanks, I don't think this is the problem, I just realised a
potential flaw in my design something very different, I will post anew.