Your manipulator should be declared as a function which accepts just one argument of type ostream&
. However, if you make it a member function, you know there is an implicit this
argument being passed to the function as well.
Thus, you should rather declare your manipulator as a free, non-member function, making it friend
of your class so that it can access its private member ib
:
class IndentStream : public ostream {
public:
IndentStream(ostream &os) : ib(os.rdbuf()), ostream(&ib){};
ostream& indent(ostream& stream) {
ib.indent();
return stream;
}
friend ostream& deindent(ostream& stream);
// ^^^^^^
private:
indentbuf ib;
};
ostream& deindent(ostream& stream)
{
IndentStream* pIndentStream = dynamic_cast<IndentStream*>(&stream);
if (pIndentStream != nullptr)
{
pIndentStream->ib.deindent();
}
return stream;
}
int main()
{
IndentStream is(cout);
is << "31 hexadecimal: " << hex << 31 << endl;
is << "31 hexadecimal: " << hex << 31 << deindent << endl;
is << "31 hexadecimal: " << hex << 31 << endl;
return 0;
}
Alternatively, if you really want your function to be a member, you could make it static:
class IndentStream : public ostream {
public:
IndentStream(ostream &os) : ib(os.rdbuf()), ostream(&ib){};
ostream& indent(ostream& stream) {
ib.indent();
return stream;
}
static ostream& deindent(ostream& stream)
{
IndentStream* pIndentStream = dynamic_cast<IndentStream*>(&stream);
if (pIndentStream != nullptr)
{
pIndentStream->ib.deindent();
}
return stream;
}
private:
indentbuf ib;
};
However, this would force you to use a qualified name to refer to it:
int main()
{
IndentStream is(cout);
is << "31 hexadecimal: " << hex << 31 << endl;
is << "31 hexadecimal: " << hex << 31 << IndentStream::deindent << endl;
// ^^^^^^^^^^^^^^
is << "31 hexadecimal: " << hex << 31 << endl;
return 0;
}