help: reference to a pointer

[xuatla]

Hi,

I have a class

class myType
{
private:
int size;
double *elem;
.....
};

now I want to get access to "elem" directly outside the class.

I defined the following member functions:

(1) double* getele() { return elem; };

(2) double*& getele() { return elem; };

And I have a function as following:
void func( int n, double *v )
{
change v;
}

myType t;
func( n, t.getele() );

When I called func in above both ways of getele, "elem" are changed. I
think the reason is because "elem" is a pointer itself. Am I right?

But do I need to use reference (2) here? Is it better in any reasons?

Thanks for your instruction.

xuatla

[Victor Bazarov]

xuatla wrote:

I have a class

class myType
{
private:
int size;
double *elem;
....
};

now I want to get access to "elem" directly outside the class.
That's a rather bad idea.

I defined the following member functions:

(1) double* getele() { return elem; };

(2) double*& getele() { return elem; };
Why two of them? Or are they only two variations that you're testing?

And I have a function as following:
void func( int n, double *v )
{
change v;
Change 'v' in what way? 'v' is passed by value, changing the 'v' itself
is rather pointless here. Are you trying to change what 'v' points to?
}

myType t;
func( n, t.getele() );

When I called func in above both ways of getele, "elem" are changed. I
think the reason is because "elem" is a pointer itself. Am I right?
What? I have no idea what you meant by this paragraph.

But do I need to use reference (2) here? Is it better in any reasons?

Judging from the 'func' implementation, no, you don't need a reference.

Victor

[xuatla]

sorry for my bad expression ...

I have wrote some in a reply to my another post.

Victor Bazarov wrote:

xuatla wrote:

I have a class

class myType
{
private:
int size;
double *elem;
....
};

now I want to get access to "elem" directly outside the class.

That's a rather bad idea.

I know...
I just thought it's easy. I set it as private instead of public to
protect it, but now I want to get access to it to simplify my work.
Maybe I should think about other ways.

I defined the following member functions:

(1) double* getele() { return elem; };

(2) double*& getele() { return elem; };

Why two of them? Or are they only two variations that you're testing?

Just for my test purpose. I want to know whether I have to use reference
here or not and which one is better.

What I want to do for elem is the change it after I call it from some
functions. And I think "keeping change" = "called by address or reference".

And I have a function as following:
void func( int n, double *v )
{
change v;

Change 'v' in what way? 'v' is passed by value, changing the 'v' itself
is rather pointless here. Are you trying to change what 'v' points to?

}
e.g.,
void func( int n, double *v )
{
for ( int i = 0; i < n; i++ ) v[i] = v[n-1-i];
}

v will be changed by func.
and I use func( n, t.getele() );

t.getele() is a double* and will be changed by func.

What I thought ( and is a mistake ) is that the two ways (1) double* getele() { return elem; }; t.getele() is just a "copy", passed by value, therefore after
func( n, t.getele() ), t.elem is not changed.
(2) double*& getele() { return elem; }; t.getele() is same as t.elem, therefore after func, t.elem is changed.

myType t;
func( n, t.getele() );

When I called func in above both ways of getele, "elem" are changed. I
think the reason is because "elem" is a pointer itself. Am I right?

What? I have no idea what you meant by this paragraph.

forget this. I made a mistake. restated in above. I made mistake on the
reference type in argument & return value.

But do I need to use reference (2) here? Is it better in any reasons?

Judging from the 'func' implementation, no, you don't need a reference.

Victor

Thanks a lot :) I am catching up now.

xuatla

[Victor Bazarov]

"xuatla" <xu****@gmail.com> wrote...

sorry for my bad expression ...

I have wrote some in a reply to my another post.

Victor Bazarov wrote:

xuatla wrote:

I have a class

class myType
{
private:
int size;
double *elem;
....
};

now I want to get access to "elem" directly outside the class.

That's a rather bad idea.

I know...
I just thought it's easy. I set it as private instead of public to protect
it, but now I want to get access to it to simplify my work.
Maybe I should think about other ways.

I defined the following member functions:

(1) double* getele() { return elem; };

(2) double*& getele() { return elem; };

Why two of them? Or are they only two variations that you're testing?

Just for my test purpose. I want to know whether I have to use reference
here or not and which one is better.

What I want to do for elem is the change it after I call it from some
functions. And I think "keeping change" = "called by address or
reference".

And I have a function as following:
void func( int n, double *v )
{
change v;

Change 'v' in what way? 'v' is passed by value, changing the 'v' itself
is rather pointless here. Are you trying to change what 'v' points to?

}
e.g.,
void func( int n, double *v )
{
for ( int i = 0; i < n; i++ ) v[i] = v[n-1-i];
}

v will be changed by func.

No. 'v' will not be changed. The contents of the memory pointed to by 'v'
will. Do you understand te difference?
and I use func( n, t.getele() );

t.getele() is a double* and will be changed by func.
Again, no, the 'double*' will not be changed. The 'double', to which
that pointer points, will.

What I thought ( and is a mistake ) is that the two ways (1) double* getele() { return elem; }; t.getele() is just a "copy", passed by value, therefore after
func( n, t.getele() ), t.elem is not changed.
Correct.
(2) double*& getele() { return elem; };

t.getele() is same as t.elem, therefore after func, t.elem is changed.

If you only use that 'getele()' as an r-value, then it won't change. But
if you do

t.getele() = something;

then with a reference you will be able to change the value of the pointer
(but I don't think that's what you want, anyway).

Victor

[xuatla]

Victor Bazarov wrote:

"xuatla" <xu****@gmail.com> wrote...

sorry for my bad expression ...

I have wrote some in a reply to my another post.

Victor Bazarov wrote:

xuatla wrote:
I have a class

class myType
{
private:
int size;
double *elem;
....
};

now I want to get access to "elem" directly outside the class.
That's a rather bad idea.
I know...
I just thought it's easy. I set it as private instead of public to protect
it, but now I want to get access to it to simplify my work.
Maybe I should think about other ways.

I defined the following member functions:

(1) double* getele() { return elem; };

(2) double*& getele() { return elem; };
Why two of them? Or are they only two variations that you're testing?

Just for my test purpose. I want to know whether I have to use reference
here or not and which one is better.

What I want to do for elem is the change it after I call it from some
functions. And I think "keeping change" = "called by address or
reference".

And I have a function as following:
void func( int n, double *v )
{
change v;
Change 'v' in what way? 'v' is passed by value, changing the 'v' itself
is rather pointless here. Are you trying to change what 'v' points to?
}

e.g.,
void func( int n, double *v )
{
for ( int i = 0; i < n; i++ ) v[i] = v[n-1-i];
}

v will be changed by func.

No. 'v' will not be changed. The contents of the memory pointed to by 'v'
will. Do you understand te difference?

O, I got it.
What I want is to change the "content". I misused the name of the
"pointer value" and "pointed value".
Thanks :-)

and I use func( n, t.getele() );

t.getele() is a double* and will be changed by func.

Again, no, the 'double*' will not be changed. The 'double', to which
that pointer points, will.

What I thought ( and is a mistake ) is that the two ways

(1) double* getele() { return elem; };

t.getele() is just a "copy", passed by value, therefore after
func( n, t.getele() ), t.elem is not changed.

Correct.

(2) double*& getele() { return elem; };

t.getele() is same as t.elem, therefore after func, t.elem is changed.

If you only use that 'getele()' as an r-value, then it won't change. But
if you do

t.getele() = something;

then with a reference you will be able to change the value of the pointer
(but I don't think that's what you want, anyway).

Now I see. I don't need to change the address of elem, just the value.

xuatla
Victor