Re: reference to a pointer
Alf P. Steinbach /Usenet wrote:
I'm a client of a
function that requires it.
That's vague, to say the least.
I interpreted it to say that he needs to call a function that requires
an argument of type toto*& (from inside a function that only has access
to a toto&).
Can you tell me how I can shorten the initalisation of totoToRef
below? I'm obviously going the long way about it.
void mymethod(toto& totoRef)
{
toto* ptr =&totoRef;
toto*& totoRefToPointer = ptr;
}
There is no `totoToRef` (which you're asking about) above.
`totoRefToPointer` is identically the same as `totoRef`.
I didn't understand this affirmation.
Assume
struct toto {
void foo() {}
}
then it seems that the two following would be legal in mymethod:
totoRef.foo();
totoRefToPointer->foo();
Right? If so, how can totoRef and totoRefToPointer be "identically the
same"? (BTW, what's the difference between "identical to", "the same as"
and "identically the same as"?)
This code jumps through hoops to do nothing.
It seems to create a reference to a pointer to an object of which only a
reference to it is accessible.
Gerhard