difference between return &*i and return i;

[Ganesh Gella]

Hi All,

At some places in code, I see following statements,

const XMLCh* some_func()
{
return &*(*i)->m_Value.begin();
}

Is that any different from following syntax ?

const XMLCh* some_func()
{
return (*i)->m_value.begin();
}

Thanks
Ganesh.

 

[Andrey Tarasevich]

At some places in code, I see following statements,

const XMLCh* some_func()
{
return &*(*i)->m_Value.begin();
}

Is that any different from following syntax ?

const XMLCh* some_func()
{
return (*i)->m_value.begin();
}
...

'i' could easily be different from '&*i' if the '*' operator is
overloaded in 'i'. That appears to be the case in the code above.

 

[Ron Natalie]

Hi All,

At some places in code, I see following statements,

const XMLCh* some_func()
{
return &*(*i)->m_Value.begin();
}

If the value returned by begin() has an overload for operator*
(or the result of *begin() has an overload for &), it's not the
same as omitting the &*.

For example if m_Value is list<XMLCh> type, then begin() returns
list<XLMCh>::iterator which is NOT the same as XMLCh*. It does
however have an operator* that reutrns XMLCh& and you can take the
address of that to get the adderss of the XMLCh object in the list.

 

[John Harrison]

Hi All,

At some places in code, I see following statements,

const XMLCh* some_func()
{
return &*(*i)->m_Value.begin();
}

Is that any different from following syntax ?

const XMLCh* some_func()
{
return (*i)->m_value.begin();
}

Thanks
Ganesh.

Yes, the return from begin() is (presumably) an iterator, and the return
from &* ... begin() is (presumably) a pointer.

john

 

[Stuart Gerchick]

Hi All,

At some places in code, I see following statements,

const XMLCh* some_func()
{
return &*(*i)->m_Value.begin();
}

Is that any different from following syntax ?

const XMLCh* some_func()
{
return (*i)->m_value.begin();
}

Thanks
Ganesh.

outside of operator loading, yes. but with operator overloading it can be anything