Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; // OK
Bar *b; // OK
Bar c; // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
private variable
6  2215 
fl wrote:
Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; // OK
Bar *b; // OK
Bar c; // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
Passing it as a parameter to the constructor is one way. this (the instance
pointer) is another. In fact, class member references have to be
initialized in the constructor initialization list (I know of no other way)
and passing as a paramter would be the usuall way. Something like (untested
code)
class Bar {
public:
Bar( Bar& foo ): d( foo ) {}
private:
Bar& d;
};
--
Jim Langston ta*******@rocke tmail.com
On Dec 30, 11:24 pm, fl <rxjw...@gmail. comwrote:
Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; // OK
Bar *b; // OK
Bar c; // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
private variable
the original object, in a special case like this one, would have to
refer to itself, which would then basicly mean that such a class could
not have a static member since the static member has no 'this'.
Its a special case, don't dissmiss references. They solve many, many
problems.
[10.7] Should you use the this pointer in the constructor? http://www.parashift.com/c++-faq-lit....html#faq-10.7
#include <iostream>
class A
{
const A& r_a;
public:
A() : r_a(*this) { }
A(const A& copy) : r_a(copy) { }
A& operator=(const A& rhv); // disabled
A const& get_r() const { return r_a; }
};
void foo(const A& r)
{
std::cout << "&r = " << &r;
std::cout << "\tr.r_a = " << &r.get_r();
std::cout << std::endl;
}
int main()
{
A a;
foo(a);
A another = a; // is NOT an assignment
foo(another);
}
/*
&r = 0x7fff0f2e1930 r.r_a = 0x7fff0f2e1930
&r = 0x7fff0f2e1920 r.r_a = 0x7fff0f2e1930
*/
On 31 déc, 04:27, Salt_Peter <pj_h...@yahoo. comwrote:
On Dec 30, 11:24 pm, fl <rxjw...@gmail. comwrote:
Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; * * *// OK
Bar *b; * * * * * * *// OK
Bar c; * * * * * * * // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
private variable
the original object, in a special case like this one, would have to
refer to itself, which would then basicly mean that such a class could
not have a static member since the static member has no 'this'.
Its a special case, don't dissmiss references. They solve many, many
problems.
[10.7] Should you use the this pointer in the constructor?http://www.parashift.com/c++-faq-lit....html#faq-10.7
#include <iostream>
class A
{
* const A& r_a;
public:
* A() : r_a(*this) { }
* A(const A& copy) : r_a(copy) { }
* A& operator=(const A& rhv); // disabled
* A const& get_r() const { return r_a; }
};
void foo(const A& r)
{
* std::cout << "&r = " << &r;
* std::cout << "\tr.r_a = " << &r.get_r();
* std::cout << std::endl;
}
int main()
{
* A a;
* foo(a);
* A another = a; // is NOT an assignment
* foo(another);
}
/*
&r = 0x7fff0f2e1930 * * r.r_a = 0x7fff0f2e1930
&r = 0x7fff0f2e1920 * * r.r_a = 0x7fff0f2e1930
*/- Masquer le texte des messages précédents -
- Afficher le texte des messages précédents -
Hi,
I find the modified code,see below, has the same output as yours.
Why the overload:
A another = a; // is NOT an assignment
does not take effect? Thank you very much.
---------------------
#include <iostream>
class A
{
const A& r_a;
public:
A() : r_a(*this) { }
// A(const A& copy) : r_a(copy) { }
// A& operator=(const A& rhv); // disabled
A const& get_r() const { return r_a; }
};
void foo(const A& r)
{
std::cout << "&r = " << &r;
std::cout << "\tr.r_a = " << &r.get_r();
std::cout << std::endl;
}
int main()
{
A a;
foo(a);
A another; // = a; // is NOT an assignment
foo(another);
}
On 31 déc, 04:27, Salt_Peter <pj_h...@yahoo. comwrote:
On Dec 30, 11:24 pm, fl <rxjw...@gmail. comwrote:
Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; * * *// OK
Bar *b; * * * * * * *// OK
Bar c; * * * * * * * // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
private variable
the original object, in a special case like this one, would have to
refer to itself, which would then basicly mean that such a class could
not have a static member since the static member has no 'this'.
Its a special case, don't dissmiss references. They solve many, many
problems.
[10.7] Should you use the this pointer in the constructor?http://www.parashift.com/c++-faq-lit....html#faq-10.7
#include <iostream>
class A
{
* const A& r_a;
public:
* A() : r_a(*this) { }
* A(const A& copy) : r_a(copy) { }
* A& operator=(const A& rhv); // disabled
* A const& get_r() const { return r_a; }
};
void foo(const A& r)
{
* std::cout << "&r = " << &r;
* std::cout << "\tr.r_a = " << &r.get_r();
* std::cout << std::endl;
}
int main()
{
* A a;
* foo(a);
* A another = a; // is NOT an assignment
* foo(another);
}
/*
&r = 0x7fff0f2e1930 * * r.r_a = 0x7fff0f2e1930
&r = 0x7fff0f2e1920 * * r.r_a = 0x7fff0f2e1930
*/- Masquer le texte des messages précédents -
- Afficher le texte des messages précédents -
Sorry, my previous response is not right, i.e. Your's
A another = a; // is NOT an assignment
is right.
My question now is: why there is no effect when I comment out:
// A(const A& copy) : r_a(copy) { }
// A& operator=(const A& rhv); // disabled
Maybe the system does the copy in a special initializer fashion?
I know now the object of "A" class is an address, which points to
itself. That is from the following two lines. Right?
const A& r_a;
public:
A() : r_a(*this) { }
---------------
Then, what's the meaning of "another" after "another=a" ?
I am not even clear:
A(const A& copy) : r_a(copy) { }
Thanks.
fl <rx*****@gmail. comwrote:
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; // OK
Bar *b; // OK
Bar c; // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
I have yet to see why one should have a non-static reference in a class
in any case, no matter what it refers to.
On Dec 31, 10:35 am, fl <rxjw...@gmail. comwrote:
On 31 déc, 04:27, Salt_Peter <pj_h...@yahoo. comwrote:
On Dec 30, 11:24 pm, fl <rxjw...@gmail. comwrote:
Hi,
There is a question about nonstatic member. C++ primer says: A
nonstatic member is restricted to being declared as a pointer or
reference to an object of its class. It only gives an example of
pointer *b.
class Bar {
public:
private:
static Bar a; // OK
Bar *b; // OK
Bar c; // error
My question is how a nonstatic member is declared as a reference to an
object of its class. Because a reference is legal only after the
original variable has been declared, where is the original object? I
feel it is really bizarre. Could you give me an example? Thanks in
advance.
private variable
the original object, in a special case like this one, would have to
refer to itself, which would then basicly mean that such a class could
not have a static member since the static member has no 'this'.
Its a special case, don't dissmiss references. They solve many, many
problems.
[10.7] Should you use the this pointer in the constructor?http://www.parashift.com/c++-faq-lit....html#faq-10.7
#include <iostream>
class A
{
const A& r_a;
public:
A() : r_a(*this) { }
A(const A& copy) : r_a(copy) { }
A& operator=(const A& rhv); // disabled
A const& get_r() const { return r_a; }
};
void foo(const A& r)
{
std::cout << "&r = " << &r;
std::cout << "\tr.r_a = " << &r.get_r();
std::cout << std::endl;
}
int main()
{
A a;
foo(a);
A another = a; // is NOT an assignment
foo(another);
}
/*
&r = 0x7fff0f2e1930 r.r_a = 0x7fff0f2e1930
&r = 0x7fff0f2e1920 r.r_a = 0x7fff0f2e1930
*/- Masquer le texte des messages précédents -
- Afficher le texte des messages précédents -
Sorry, my previous response is not right, i.e. Your's
A another = a; // is NOT an assignment
is right.
My question now is: why there is no effect when I comment out:
// A(const A& copy) : r_a(copy) { }
// A& operator=(const A& rhv); // disabled
the compiler generates the copy ctor if you don't. it probably does
the exact same as the one commented out.
I prefer declaring it in cases i want to diagnose problems and assert
theories.
A(const A& copy) : r_a(copy) { std::cout << "copy A"; }
is nice to have when observing and troubleshooting .
>
Maybe the system does the copy in a special initializer fashion?
not at all. Its a member-wise copy, nothing special.
Its the same with something like
struct K
{
int n;
double d;
char c;
};
You can initialize an instance of K and you can copy it because the
compiler generates a copy ctor for you. It also generates a defult
ctor and an assignment operator (if needed).
I know now the object of "A" class is an address, which points to
itself. That is from the following two lines. Right?
const A& r_a;
public:
A() : r_a(*this) { }
---------------
Then, what's the meaning of "another" after "another=a" ?
I am not even clear:
A(const A& copy) : r_a(copy) { }
Thanks.
instance 'another' is a psuedo-copy of a. The difference is that
another's member reference doesn't refer to another.
This isn't something you need not worry about, you'll not find such a
strategy in code out there.
You do need to understand ctor, copy ctor and init list. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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