Re: why does this call the destructor?
On 4 Maj, 08:43, "michael" <s...@begone.net> wrote:
"anon" <a...@no.no> wrote in message
news:f1ejbt$t7s$2@el-srv04-CHE.srvnet.eastlink.de...
michael wrote:
ostream& operator <<(ostream& lhs, someClass rhs){
lhs << rhs.str; // This results in a call to the destructor
for the someClass object....why?
return lhs;
}
It is not that line that calls the destructor, but 2 lines later ( the }
bracket). The destruction of that object happens, because temporary obj=
ect
rhs of type SomeClass needs to be destroyed. If you used a reference as=
a
parameter, it wouldn't happen:
ostream& operator <<(ostream& lhs, const someClass& rhs)
^^^^^ ^
^^^^^ ^
Thanks for the reply, and you are of course correct the temporary rhs nee=
ds
to be destroyed hence the call, but if it is only the temporary rhs that
gets destroyed, why if I make a second << call like:
int main(){
someClass soc;
std::cout << soc;
std::cout << soc;
}
does the someClass object no longer exist?
Surely the copy should have been destroyed, not the original, especially
since I didn't pass it by reference.
The someObject still exists but you have deleted the string which the
member str points to. Take this as a lesson, when you allocate memory
in the constructor you should also make sure to implement the copy-
constructor and assignment operator (the rule of three I think it's
called) or things like this will bite you.
--
Erik Wikstr=F6m