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How to invoke member function over by "const"?

#include <string>
#include <iostream>
using namespace std;

class TConst
{
private:
string con;
string uncon;
public:
TConst():con("const"),uncon("Un const"){}
string GetString()const;
string& GetString();
};
//how can I invoke this member function?
string TConst::GetString()const
{
return con;
}
string& TConst::GetString()
{
return uncon;
}

int main()
{
TConst tc;
const string& csr=tc.GetString();
const string cs=tc.GetString();
string& sr=tc.GetString();
string s=tc.GetString();

cout<<"const string& csr="<<csr<<endl;
cout<<"const string cs="<<cs<<endl;
cout<<"string& sr="<<sr<<endl;
cout<<"string s="<<s<<endl;
}

===output=============
const string& csr=Un const
const string cs=Un const
string& sr=Un const
string s=Un const
===output end=====

why all the call "GetString()" invoke the un const function.
--
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--------------------------
Aug 3 '06 #1
16 2072
recover wrote:
#include <string>
#include <iostream>
using namespace std;

class TConst
{
private:
string con;
string uncon;
public:
TConst():con("const"),uncon("Un const"){}
string GetString()const;
string& GetString();
};
//how can I invoke this member function?
You need a const object for that.
string TConst::GetString()const
{
return con;
}
string& TConst::GetString()
{
return uncon;
}

int main()
{
TConst tc;
The 'tc' is a NON-const object (in this scope, anyway).
const string& csr=tc.GetString();
const string cs=tc.GetString();
string& sr=tc.GetString();
string s=tc.GetString();

cout<<"const string& csr="<<csr<<endl;
cout<<"const string cs="<<cs<<endl;
cout<<"string& sr="<<sr<<endl;
cout<<"string s="<<s<<endl;
}

===output=============
const string& csr=Un const
const string cs=Un const
string& sr=Un const
string s=Un const
===output end=====

why all the call "GetString()" invoke the un const function.
Because all invocations are for the 'tc' object an it is non-const.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Aug 3 '06 #2
recover wrote:
#include <string>
#include <iostream>
using namespace std;

class TConst
{
private:
string con;
string uncon;
public:
TConst():con("const"),uncon("Un const"){}
string GetString()const;
string& GetString();
};
//how can I invoke this member function?
string TConst::GetString()const
{
return con;
}
string& TConst::GetString()
{
return uncon;
}

int main()
{
TConst tc;
const string& csr=tc.GetString();
const string cs=tc.GetString();
string& sr=tc.GetString();
string s=tc.GetString();

cout<<"const string& csr="<<csr<<endl;
cout<<"const string cs="<<cs<<endl;
cout<<"string& sr="<<sr<<endl;
cout<<"string s="<<s<<endl;
}

===output=============
const string& csr=Un const
const string cs=Un const
string& sr=Un const
string s=Un const
===output end=====

why all the call "GetString()" invoke the un const function.
Because the function that is called depends upon the object making the
call (tc) and does not depend upon what you assign the result of the
function call to. If you want to call the const member function then
you need to invoke GetString on a const object.

const TConst tc;
tc.GetString();

Or, you can cast a non-const TConst to a const TConst.
Aug 3 '06 #3
recover posted:
why all the call "GetString()" invoke the un const function.

class MyClass {
public:

void Func() {}

void Func() const {}
};

int main()
{
MyClass obj;
MyClass const cobj;

obj.Func() /* Non-const version */
cobj.Func() /* Const version */

const_cast<MyClass const&>(obj).Func() /* Const version */
/* Or if you have a phobia of "const_cast"
(as many people here seem to have), then
you can use a reference instead: */

MyClass const &cr = obj;

cr.Func(); /* Const Version */
}
I wonder would it be UB to call the non-const function on a const object
(if the non-const function doesn't alter any member data)? Something like:

const_cast<MyClass&>(cobj).Func();

Anyone know if that's forbidden?

--

Frederick Gotham
Aug 3 '06 #4
Frederick Gotham wrote:
...
I wonder would it be UB to call the non-const function on a const object
(if the non-const function doesn't alter any member data)? Something like:

const_cast<MyClass&>(cobj).Func();

Anyone know if that's forbidden?
...
No, it is not UB. As long as (as you said) the member function doesn't alter any
member data

--
Best regards,
Andrey Tarasevich
Aug 3 '06 #5
Andrey Tarasevich wrote:
Frederick Gotham wrote:
>...
I wonder would it be UB to call the non-const function on a const
object (if the non-const function doesn't alter any member data)?
Something like:

const_cast<MyClass&>(cobj).Func();

Anyone know if that's forbidden?
...

No, it is not UB. As long as (as you said) the member function
doesn't alter any member data
Why is the function that doesn't alter any data *not* declared 'const',
that's what I'd be asking...

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Aug 3 '06 #6
Victor Bazarov posted:
Why is the function that doesn't alter any data *not* declared 'const',
that's what I'd be asking...

Instead of altering member objects, it could alter resources:

class MyClass {

int *const p;

MyClass() : p(new int[64]) {}

void SetElement(unsigned const index, int const val) /* Non-const */
{
p[index] = val;
}
};
Depending on the design of the class, it might make perfect sense to make the
above method non-const, even though it doesn't alter any member objects.

--

Frederick Gotham
Aug 3 '06 #7
Frederick Gotham wrote:
Victor Bazarov posted:
>Why is the function that doesn't alter any data *not* declared
'const', that's what I'd be asking...


Instead of altering member objects, it could alter resources:

class MyClass {

int *const p;

MyClass() : p(new int[64]) {}

void SetElement(unsigned const index, int const val) /* Non-const
*/ {
p[index] = val;
}
};
Depending on the design of the class, it might make perfect sense to
make the above method non-const, even though it doesn't alter any
member objects.
OK. I'll bite.

If such function is non-const, why would somebody try to call it in the
context where the object is const and thus requires a const_cast to call
the non-const member function? If the design of the class is done right,
there would be no circumstances under which a call to 'SetElement' needs
to be done in a function where 'MyClass' object is const, do you agree?

Generally speaking, it's not UB (what you asked about), but makes no
sense in a well-designed program.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Aug 3 '06 #8
Frederick Gotham wrote:
Victor Bazarov posted:
Why is the function that doesn't alter any data *not* declared 'const',
that's what I'd be asking...


Instead of altering member objects, it could alter resources:

class MyClass {

int *const p;

MyClass() : p(new int[64]) {}

void SetElement(unsigned const index, int const val) /* Non-const */
{
p[index] = val;
}
};
Depending on the design of the class, it might make perfect sense to make the
above method non-const, even though it doesn't alter any member objects.
In the spirit of making as much const as possible, I prefer using
wrappers and containers (e.g., std::vector) that enforce "deep
constness" for their wrapped/contained values and applying mutable
where that behavior is not desirable. Thus, your SetElement() function
wouldn't be declared const, not just because it shouldn't be for
semantic purposes, but because it *couldn't* be.

Cheers! --M

Aug 3 '06 #9
Victor Bazarov posted:
If such function is non-const, why would somebody try to call it in the
context where the object is const and thus requires a const_cast to call
the non-const member function? If the design of the class is done right,
there would be no circumstances under which a call to 'SetElement' needs
to be done in a function where 'MyClass' object is const, do you agree?

Generally speaking, it's not UB (what you asked about), but makes no
sense in a well-designed program.

Yes, I see what your getting at. But there are times when the programmer
has to do some "funky stuff" to achieve something which can't be achieved
via domestic means.

Let's say that there's ONE instance in our program where we want to invoke
this non-const method upon a const object; If we were to be perfectly
politically correct about it, then we would re-engineer our design.

But... then we might think, "Is it warranted to dump our brilliant design
just because we circumvent it in one measly place?".

My answer is "No". It's great for a design to be consistent, and durable,
and all that... but in my opinion, there's nothing wrong with "digging
under the fence" once or twice -- so long as it all comes together
perfectly in the end.

Think back to when you first encountered "const_cast". I think EVERY
person's first reaction was, "Oh, this has GOT to be dirty; what could it
possibly achieve legitimately?". Over time, we began to see that
"const_cast" really isn't dirty at all -- it just provides us a way of
"digging under the fence".

So, depending on what our program does, and depending on what our class
does, there may be legitimate places throughout the program where we invoke
this non-const method upon a const object... ?

--

Frederick Gotham
Aug 3 '06 #10
mlimber posted:
>Instead of altering member objects, it could alter resources:

class MyClass {

int *const p;

MyClass() : p(new int[64]) {}

void SetElement(unsigned const index, int const val) /* Non-const
*/ {
p[index] = val;
}
};
Depending on the design of the class, it might make perfect sense to
make the above method non-const, even though it doesn't alter any
member objects.
In the spirit of making as much const as possible, I prefer using
wrappers and containers (e.g., std::vector) that enforce "deep
constness" for their wrapped/contained values and applying mutable
where that behavior is not desirable. Thus, your SetElement() function
wouldn't be declared const, not just because it shouldn't be for
semantic purposes, but because it *couldn't* be.

Not sure what you're getting at... ? In the above code, "SetElement" could
certainly be defined as "const", as it doesn't alter any member objects.

--

Frederick Gotham
Aug 3 '06 #11
Frederick Gotham <fg*******@spam.comwrote:
mlimber posted:
>Frederick Gotham wrote:
>>Instead of altering member objects, it could alter resources:

class MyClass {

int *const p;

MyClass() : p(new int[64]) {}

void SetElement(unsigned const index, int const val) /* Non-const
*/ {
p[index] = val;
}
};
Depending on the design of the class, it might make perfect sense to
make the above method non-const, even though it doesn't alter any
member objects.
>In the spirit of making as much const as possible, I prefer using
wrappers and containers (e.g., std::vector) that enforce "deep
constness" for their wrapped/contained values and applying mutable
where that behavior is not desirable. Thus, your SetElement() function
wouldn't be declared const, not just because it shouldn't be for
semantic purposes, but because it *couldn't* be.

Not sure what you're getting at... ? In the above code, "SetElement" could
certainly be defined as "const", as it doesn't alter any member objects.
What he said was:
>In the spirit of making as much const as possible, I prefer using
wrappers and containers (e.g., std::vector) that enforce "deep
constness" for their wrapped/contained values
In other words, instead of using a raw pointer, he would prefer using
e.g. a std::vector<int>.

class MyClass {
std::vector<intp;

MyClass() : p(64) {}

void SetElement(unsigned const index, int const val)
{
p[index] = val;
}
};

In this situation, you can not declare SetElement as const unless you
make p mutable.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply
Aug 3 '06 #12
Frederick Gotham wrote:
Victor Bazarov posted:
>If such function is non-const, why would somebody try to call it in
the context where the object is const and thus requires a const_cast
to call the non-const member function? If the design of the class
is done right, there would be no circumstances under which a call to
'SetElement' needs to be done in a function where 'MyClass' object
is const, do you agree?

Generally speaking, it's not UB (what you asked about), but makes no
sense in a well-designed program.


Yes, I see what your getting at. But there are times when the
programmer has to do some "funky stuff" to achieve something which
can't be achieved via domestic means.
In the world where design rules, there could not be such times.
Let's say that there's ONE instance in our program where we want to
invoke this non-const method upon a const object; If we were to be
perfectly politically correct about it, then we would re-engineer our
design.
Yes.
But... then we might think, "Is it warranted to dump our brilliant
design just because we circumvent it in one measly place?".
If our brilliant design allows for such situation to occur, it's not
brilliant.
My answer is "No". It's great for a design to be consistent, and
durable, and all that... but in my opinion, there's nothing wrong
with "digging under the fence" once or twice -- so long as it all
comes together perfectly in the end.
Here you go, you already used the sacramental "if". If you dig under
the fence, how can you be sure that "all comes together perfectly"?
I submit that you cannot.
Think back to when you first encountered "const_cast". I think EVERY
person's first reaction was, "Oh, this has GOT to be dirty; what
could it possibly achieve legitimately?". Over time, we began to see
that "const_cast" really isn't dirty at all -- it just provides us a
way of "digging under the fence".
Hey, if it's in the language, it must be useful for something, right?
So, depending on what our program does, and depending on what our
class does, there may be legitimate places throughout the program
where we invoke this non-const method upon a const object... ?
I've not seen _one_ such place. Honestly.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Aug 3 '06 #13
Frederick Gotham wrote:
>
Yes, I see what your getting at. But there are times when the programmer
has to do some "funky stuff" to achieve something which can't be achieved
via domestic means.
You're saying I can't always program from home? ;)

Aug 3 '06 #14
Old Wolf wrote:
Frederick Gotham wrote:

Yes, I see what your getting at. But there are times when the
programmer has to do some "funky stuff" to achieve something which
can't be achieved via domestic means.

You're saying I can't always program from home? ;)
You gotta put pants on sometimes.


Brian
Aug 4 '06 #15
Default User wrote:
Old Wolf wrote:
>Frederick Gotham wrote:
>>Yes, I see what your getting at. But there are times when the
programmer has to do some "funky stuff" to achieve something which
can't be achieved via domestic means.
You're saying I can't always program from home? ;)

You gotta put pants on sometimes.
Sez you!!!!
Aug 4 '06 #16
Victor Bazarov wrote:
>Frederick Gotham wrote:
>So, depending on what our program does, and depending on what our
class does, there may be legitimate places throughout the program
where we invoke this non-const method upon a const object... ?

I've not seen _one_ such place. Honestly.
While I agree with this in practice, there are times when I've used
const_cast just to avoid having the same functionality in multiple
places.

E.g.:

class MyClass
{
int& operator[](int i);
const int& operator[](int i) const;
};

If the calculations to compute what to return are "complicated," it
could be that one of these functions may want to const_cast in and out
of constness for a call to the other version. There is no way that I
know of to do this without const_cast unless you've got an
internal/private method which is const but returns a non-const
reference, but then you're breaking other rules anyway (and would it
even compile?)...

Aug 5 '06 #17

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