Reference of the return value.

[sss.zhou]

Question about reference of the function return object.

class A
{
public:
std::string getString() const {
return str_;
}

private:
std:: string str_;
};

function_ref(const A& a){
...
const std::string& str = a.getString(); //getString ()
return 'std::string' not 'const std:string&',
cout << "Text is: " << str; //use str here will be OK
...
}

function_point(const A& a) {
...
const std::string* pStr = &a.getString(); //the return
object of 'std::string' will destruct here
cout << "Text is: " << *pStr; //so using '*pStr' here may
cause a crash
...
}

My question is if there had any problem with reference of the
function return object, because 'a.getString()' just return a temp
object.
After I debugged this in VS2003, I found that the temp object will
be destroyed at the end of this function.
'function_ref(const A& a)' will be running without any problem in
VS2003 but 'function_point(const A& a)' may cause error.

I want to ask if that reference of a temp return value delay the
destruction is just a property of MS VS2003 or C++ standard.

[Ron Natalie]

ss******@gmail.com wrote:

My question is if there had any problem with reference of the
function return object, because 'a.getString()' just return a temp
object.

No. The return is a temporary rvalue. If you bind it to a const
reference it will live as long as the reference.

Pointers, however, have no inteligence about what they refer to.
The temporary rvalue from the return, when not bound to a reference,
is destroyed at the end of the full expression it is created in.
In this case, it's history as soon as the initialization of the
pointer takes place.

[sss.zhou]

No. The return is a temporary rvalue. If you bind it to a const

reference it will live as long as the reference.

Pointers, however, have no inteligence about what they refer to.
The temporary rvalue from the return, when not bound to a reference,
is destroyed at the end of the full expression it is created in.
In this case, it's history as soon as the initialization of the
pointer takes place.

Thanks for you help.

You answers let me know a concept of lvalue and rvalue.

So const reference to a temporary rvalue can be using a concept of
the c++ standard, not the VS2003 property, and this is another
difference between the reference and the pointer.

[Juha Nieminen]

Ron Natalie wrote:

No. The return is a temporary rvalue. If you bind it to a const
reference it will live as long as the reference.

Is this also true for const references as member variables? If I try
this, it compiles without warnings but causes a segmentation fault
(at least with gcc 3.3):

#include <iostream>
#include <string>

class A
{
const std::string& s;

public:
A(): s("test2") {}
void print() { std::cout << s << std::endl; }
};

int main()
{
A a;
a.print();
}

[BobR]

Juha Nieminen <no****@thanks.invalidwrote in message...

Ron Natalie wrote:

No. The return is a temporary rvalue. If you bind it to a const
reference it will live as long as the reference.

Is this also true for const references as member variables? If I try
this, it compiles without warnings but causes a segmentation fault
(at least with gcc 3.3):

#include <iostream>
#include <string>

class A{
const std::string& s;
public:
A(): s("test2") {}
void print() { std::cout << s << std::endl; }
};

int main(){
A a;
a.print();
}

Think about what you are initializing the reference to.

std::string &rs( "Hi There" );
// Error: could not convert `"Hi There"' to `std::string&'
class A{
std::string s;
const std::string &rs;
public:
A(): s( "test2" ), rs( s ){}
void print() { std::cout << rs << std::endl;}
};

int main(){
A a;
a.print();
}

--
Bob R
POVrookie

[BobR]

BobR wrote in message...

>
Juha Nieminen <no****@thanks.invalidwrote in message...

Ron Natalie wrote:

No. The return is a temporary rvalue. If you bind it to a const
reference it will live as long as the reference.

Is this also true for const references as member variables? If I try
this, it compiles without warnings but causes a segmentation fault
(at least with gcc 3.3):

#include <iostream>
#include <string>
class A{
const std::string& s;
public:
A(): s("test2") {}
void print() { std::cout << s << std::endl; }
};

int main(){
A a;
a.print();
}

Think about what you are initializing the reference to.

std::string &rs( "Hi There" );
// Error: could not convert `"Hi There"' to `std::string&'

class A{
std::string s;
const std::string &rs;
public:
A(): s( "test2" ), rs( s ){}
void print() { std::cout << rs << std::endl;}
};

int main(){
A a;
a.print();
}

Oooops, sorry Juha. I see what you were doing now.
I forgot Murphy's [net] Laws - Never post while on your *first* cup of
coffee!

Oh well, maybe my example will help some newbie.
--
Bob R
POVrookie

[Gianni Mariani]

Juha Nieminen wrote:

Ron Natalie wrote:

>No. The return is a temporary rvalue. If you bind it to a const
reference it will live as long as the reference.

Is this also true for const references as member variables? If I try
this, it compiles without warnings but causes a segmentation fault
(at least with gcc 3.3):

No.

Easy test code attached.
#include <iostream>
template <int N>
struct X
{
X()
{
std::cout << "X<" << N << ">::X()\n";
}

~X()
{
std::cout << "X<" << N << ">::~X()\n";
}
};
struct A
{
const X<1& x;

A()
: x( X<1>() )
{
std::cout << "A::A()\n";
}

~A()
{
std::cout << "A::~A()\n";
}
};
int main()
{
const X<2& x2 = X<2>();
A a;
}

[Juha Nieminen]

Gianni Mariani wrote:

Juha Nieminen wrote:

> Is this also true for const references as member variables?

No.

Why doesn't gcc issue at least a warning? It's assigning a temporary
to the reference. Can't it see that in this case it will cause misbehavior?

[Gianni Mariani]

Juha Nieminen wrote:

Gianni Mariani wrote:

>Juha Nieminen wrote:

>> Is this also true for const references as member variables?

>No.

Why doesn't gcc issue at least a warning? It's assigning a temporary
to the reference. Can't it see that in this case it will cause misbehavior?

That's a good suggestion. I'd probably file a bug with GCC and see what
happens.