On Sunday 19 March 2006 18:45, Steve Graegert wrote:
> On 3/19/06, Shriramana Sharma <samjnaa@xxxxxxxxx> wrote:
> > Suppose there is a function f() which returns an integer and does not
> > change any value of the class it belongs to, should I declare it as:
>
> I don't understand what you mean here.
>
> > const int f();
>
> By declaring a const return type you are promising that the original
> variable (inside the function's stack frame) will not be modified.
> Because you're returning it by value, it's copied so the original
> value could never be modified via the return value. This qualifier
> has not meaning for primitive builtin types. This is different for
> user-defined types: if a function returns a class object by value as
> const, the return value of that function cannot be an lvalue (that is,
> it cannot be assigned to or otherwise modified).
It can have a meaning for built in types, too. For example you could do
something like this:
int& f(...) {....}
Through this you could actually change a value inside the class. Or cause a
segfault - depending on what you return, a member or a temp variable :)
Constructs of this kind are only usable of you pass back a reference to a
member that lives beyond the function call, of course. If you use const you
make sure that your reference is read-only.
>
> > or
> >
> > int f() const;
>
> Marks a function as const allowing it to be called by const objects
> (btw, const objects can only call const member functions). This
> construct is usually found as part of member functions declarations,
> which are known to be immutable.
>
> You probably want to choose the latter one.
>
> \Steve
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