return reference to member variable

[Billy]

Hi,

I would like to return a reference to a member variable:

class foo
{
int i;
int& geti() const;
};

int& foo::geti() const
{
return i;
}

But I get the following error:
invalid initialization of reference of type 'double&' from expression
of type 'const double'

Am I missing something? This seems like it should be doable.

-Billy

 

[Rolf Magnus]

Hi,

I would like to return a reference to a member variable:

class foo
{
int i;
int& geti() const;
};

int& foo::geti() const
{
return i;
}

But I get the following error:
invalid initialization of reference of type 'double&' from expression
of type 'const double'

Am I missing something?

Isn't that error message obvious? Your member function is const, and so
within it, i is const. You're trying to bind it to a non-const referernce,
which is of course not allowed. Make it:

const int& foo::geti() const
{
return i;
}

 

[Billy]

I was under the impression that making the function const meant:
"within this function I promise not to change the state of the object"

Since I wasn't changing the state of the object I thought it should be
ok. Clearly my understanding of making a function const was overly
naive.

Thanks, Billy

 

[Karl Heinz Buchegger]

I was under the impression that making the function const meant:
"within this function I promise not to change the state of the object"

Right.
But if you give away a reference to some object internals, then any
other lousy code *is* able to change the state of the object. Thus
the fact of handing out a reference opens the wormhole.

Since I wasn't changing the state of the object

You didn't. But you enabled other code to do it.

 

[Rolf Magnus]

I was under the impression that making the function const meant:
"within this function I promise not to change the state of the object"

"... and not to expose it in a way that makes changes to it from outside
possible".

Think of something like:

foo f;
f.geti()++;

 

[Billy]

Sorry I haven't done much programming...

Example
**********
foo f;
f.geti()++;

int& foo::geti(); // Example works
int& foo::geti() const; // Error, exposing foo via return
const int& foo::geti() const; // Error, f.geti() is const
const int& foo::geti(); // Error, f.geti() is const

Am I thinking about this right?

-Thanks, b