As I understand, a const member function is used by const object to
ensure that its instance isn't modified throughout its life. Am I
missing something..
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
}
private:
char* pch;
};
int main(){
const char str[] = "I a constant";
const A* obj = new A(*str, sizeof(str));
cout<<"obj->print() = "<<obj->print()<<end l;
return 0;
}
Wg
14  1986 
Wolfgang wrote:
As I understand, a const member function is used by const object to
ensure that its instance isn't modified throughout its life. Am I
missing something..
No, you got it right. The instance of 'A' in your code below is NOT
modified in the 'print' member function. Some memory pointed to by
the member 'pch' is modified, but that memory is not part of the 'A'
object.
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
Your program has undefined behaviour here. You're supposed to use
delete[]
since you allocate 'pch' using the array form of 'new'.
}
private:
char* pch;
};
int main(){
const char str[] = "I a constant";
const A* obj = new A(*str, sizeof(str));
Does it even compile? Your constructor takes a reference to non-const
char, and you're dereferencing a pointer to const char...
cout<<"obj->print() = "<<obj->print()<<end l;
return 0;
}
Wg
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
On Jun 15, 5:54 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
Wolfgang wrote:
As I understand, a const member function is used by const object to
ensure that its instance isn't modified throughout its life. Am I
missing something..
No, you got it right. The instance of 'A' in your code below is NOT
modified in the 'print' member function. Some memory pointed to by
the member 'pch' is modified, but that memory is not part of the 'A'
object.
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
Your program has undefined behaviour here. You're supposed to use
delete[]
If my understanding is correct, delete[] is called when we want to
delete an array of objects - destructor needs to be called for each
object.
Won't be Ok if we use " delete pch " for deleting char array ?
>
since you allocate 'pch' using the array form of 'new'.
}
private:
char* pch;
};
int main(){
const char str[] = "I a constant";
Sorry use -
char str[] = "I a constant"; // instead of const char
str[] = "I a constant";
>
const A* obj = new A(*str, sizeof(str));
Does it even compile? Your constructor takes a reference to non-const
char, and you're dereferencing a pointer to const char...
cout<<"obj->print() = "<<obj->print()<<end l;
return 0;
}
Wg
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Well see carefully what you're doing
char* print() const {
*pch = 'M';
return pch;
}
you are modifying the CONTENTS at the location pointed by pch and not
the pch
for example pch pointing to 1000 location in memory your code is
modifying the contents at 1000 not the value contained in by
'pch'(i.e. 1000).
On Jun 15, 6:26 pm, Amol <amolj.1...@gma il.comwrote:
Well see carefully what you're doing
char* print() const {
*pch = 'M';
return pch;
}
you are modifying the CONTENTS at the location pointed by pch and not
the pch
for example pch pointing to 1000 location in memory your code is
modifying the contents at 1000 not the value contained in by
'pch'(i.e. 1000).
Infact, why is it that the compiler is permitting change to the
particular text passed in ( text with which the object is
initialized). Aren't we modifying how the object has been
initialized ?
While a const object shouldn't be modified throughout its life time.
Wolfgang wrote:
On Jun 15, 6:26 pm, Amol <amolj.1...@gma il.comwrote:
>Well see carefully what you're doing
char* print() const {
>> *pch = 'M';
return pch;
}
you are modifying the CONTENTS at the location pointed by pch and not
the pch
for example pch pointing to 1000 location in memory your code is
modifying the contents at 1000 not the value contained in by
'pch'(i.e. 1000).
Infact, why is it that the compiler is permitting change to the
particular text passed in ( text with which the object is
initialized).
The text is not part of the constant object.
Aren't we modifying how the object has been
initialized ?
You're modifying something else only sideways related to your object.
It's essentially the same as you would, for example, output something
in the constant function (isn't that what 'print' is for?), like cout.
The memory you're modifying does NOT belong to your object. Your 'A'
object simply has a pointer to it. There can be countless number of
pointers to the same memory and anybody can modify it. Just like
you can access 'cout' in your function and actually modify it (by
outputting to it).
While a const object shouldn't be modified throughout its life time.
Yes. How is that relevant here?
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Wolfgang wrote:
On Jun 15, 5:54 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
~A(){
delete pch;
Your program has undefined behaviour here. You're supposed to use
delete[]
If my understanding is correct, delete[] is called when we want to
delete an array of objects - destructor needs to be called for each
object.
Yes, but also the free store management routines need to know what you are
deleting.
Won't be Ok if we use " delete pch " for deleting char array ?
No. Why would Victor tell you that it's wrong if it wasn't?
--
rbh
On Jun 15, 6:41 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
Wolfgang wrote:
On Jun 15, 6:26 pm, Amol <amolj.1...@gma il.comwrote:
Well see carefully what you're doing
char* print() const {
> *pch = 'M';
return pch;
}
you are modifying the CONTENTS at the location pointed by pch and not
the pch
for example pch pointing to 1000 location in memory your code is
modifying the contents at 1000 not the value contained in by
'pch'(i.e. 1000).
Infact, why is it that the compiler is permitting change to the
particular text passed in ( text with which the object is
initialized).
The text is not part of the constant object.
Aren't we modifying how the object has been
initialized ?
You're modifying something else only sideways related to your object.
It's essentially the same as you would, for example, output something
in the constant function (isn't that what 'print' is for?), like cout.
The memory you're modifying does NOT belong to your object. Your 'A'
object simply has a pointer to it. There can be countless number of
pointers to the same memory and anybody can modify it. Just like
you can access 'cout' in your function and actually modify it (by
outputting to it).
While a const object shouldn't be modified throughout its life time.
Yes. How is that relevant here?
V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask
Modifying the code a little bit here, esp changing the data member to
be public - so that I can access outside the class.
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
}
public:
char* pch;
};
int main(){
char str[] = "I a constant";
const A* obj = new A(*str, sizeof(str));
cout<<"obj->print() = "<<obj->print()<<end l;
cout<<"obj->pch = "<<obj->pch<<endl;
return 0;
}
I'm trying to print the member data which was initialised on the first
place, but modified on calling print().
Wg
Wolfgang wrote:
On Jun 15, 6:41 pm, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:
>Wolfgang wrote:
On Jun 15, 6:26 pm, Amol <amolj.1...@gma il.comwrote:
Well see carefully what you're doing
char* print() const {
>> *pch = 'M';
return pch;
}
>you are modifying the CONTENTS at the location pointed by pch and not
the pch
for example pch pointing to 1000 location in memory your code is
modifying the contents at 1000 not the value contained in by
'pch'(i.e. 1000).
Infact, why is it that the compiler is permitting change to the
particular text passed in ( text with which the object is
initialized).
The text is not part of the constant object.
Aren't we modifying how the object has been
initialized ?
You're modifying something else only sideways related to your object.
It's essentially the same as you would, for example, output something
in the constant function (isn't that what 'print' is for?), like cout.
The memory you're modifying does NOT belong to your object. Your 'A'
object simply has a pointer to it. There can be countless number of
pointers to the same memory and anybody can modify it. Just like
you can access 'cout' in your function and actually modify it (by
outputting to it).
While a const object shouldn't be modified throughout its life time.
Yes. How is that relevant here?
Modifying the code a little bit here, esp changing the data member to
be public - so that I can access outside the class.
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
More commonly, this should be:
A(char& str, size_t sz) : pch(new char[sz]) {
strncpy(pch, &str, sz);
}
Also: most C++ programmers would expect a C-style string passed as a const
char*, not a char&.
>
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
delete [] pch;
Even though "delete pch" would work on many compilers, it _is_ undefined
behaviour. Don't do it!
}
public:
char* pch;
};
Now, a const A would look like this:
"const class" A {
// member functions
public:
char *const pch;
};
see that pch is const, but it points to non const data.
>
int main(){
char str[] = "I a constant";
const A* obj = new A(*str, sizeof(str));
cout<<"obj->print() = "<<obj->print()<<end l;
cout<<"obj->pch = "<<obj->pch<<endl;
obj->pch = str // error -- pch is const.
return 0;
}
I'm trying to print the member data which was initialised on the first
place, but modified on calling print().
Yes. This should print "M a constant" in both lines of output. As Victor
says: The text pointed to is not part of *obj as far as C++ is concerned,
only the bits in the pointer itself is. If you want this data to appear as
a member of A, assure A's member functions treat it as such: If print
returned a const char*, and did not modify *pch, then your code would not
be able to modify the data through obj either. (Without performing a
const_cast, that is).
--
rbh
Wolfgang wrote:
[..]
Modifying the code a little bit here, esp changing the data member to
be public - so that I can access outside the class.
My comments inline with code relate only to style, not to functionality.
#include <iostream>
using namespace std;
class A{
public:
A(char& str, size_t sz){
pch = new char [sz];
strncpy(pch, &str, sz);
}
A(char const* str, size_t sz) : pch(new char[sz]) {
strncpy(pch, str, sz);
}
>
char* print() const {
*pch = 'M';
return pch;
}
~A(){
delete pch;
}
public:
char* pch;
};
int main(){
char str[] = "I a constant";
char const str[] = "I a constant";
>
const A* obj = new A(*str, sizeof(str));
const A* obj = new A(str, sizeof(str));
cout<<"obj->print() = "<<obj->print()<<end l;
cout<<"obj->pch = "<<obj->pch<<endl;
return 0;
}
I'm trying to print the member data which was initialised on the first
place, but modified on calling print().
I think you're missing something conceptual here. The data member
is _a pointer_, not the memory you allocated using 'new[]'. The
member, a pointer, does NOT change. You can print its value by
casting it to (void*). You will see that 'print' does not change
the value of the pointer and therefore does not change the value of
the constant object.
What you are changing is the contents of some memory which the 'pch'
pointer points to. That memory does NOT belong to your 'obj'. NOT.
Do you not understand that? There is no other ownership relationship
between 'obj' and the memory you allocate in 'A::A' than in your mind.
The compiler cannot prevent you from modifying that memory since it
is NOT part of the constant object 'obj'. Only _immediate_ parts of
the constant object cannot be modified, like 'pch' value itself. Try
assigning a different value to 'pch' (not to '*pch'), and you will
see what is meant by contant object.
V
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