RE: [Dev-C++] returning a constant string reference

[Ioannis V. <no...@ho...>]

> -----Original Message-----
> From: dev...@li...=20
> [mailto:dev...@li...] On Behalf=20
> Of Daniel Glenfield
> Sent: Thursday, October 17, 2002 2:20 PM
> To: dev...@li...
> Subject: [Dev-C++] returning a constant string reference
>=20
>=20
>=20
>=20
> This is the code currently
> String GetName() const;

std::string GetName() const;


This returns a (temporary) object created using the copy constructor.


So for example if you have:


class something
{
 string s;
 public:
 something(const string *n)
 string GetName() const { return s; }
};

// ...

something so("Johnson");

string l=3Dso.GetName();


At the so.GetName() call

a temporary using the (default here) copy constructor
something::something(const something &) is called which is assigned to l
using the (default here) assignment something::operator=3D(const =
something
&). After the end of the expression the temporary is usually  destroyed
(and not at the end of the scope).







>=20
> To make the function return a const string reference, do I changeit to
>=20
> String& GetName() const;
>=20
> or is it now returning a String object?


std::string &GetName();


It is returning a reference to an existing object (i assume you intend
to the data member).


Ioannis

* Ioannis Vranos
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