Question on iostream - c++

shnya <testmhigashi> wrote:

> I don't know why following code is working.

The code you posted shouldn't even compile. I've corrected the obvious
errors below:

ostream&
operator<<(ostream& os, CPerson &str){
return os << str.getName();
/* getName returns const char* */
}

int
main()
{
CPerson person; //Person is User-Definition Class

// not strange
cout << person;

ofstream ofs("test.txt");

// strange. why do this code work?
ofs << person;
}

> I read the book "Programming C++ third edition" & "The C++ Standard
> Library - A Tutorial and Reference" again.
> And I consult the source code which is g++ 4.0 on Mac OSX.
>
> ofstream is inherited ostream.
> but "operator<<" is not member operator, no virtual directive, and
> ofstream object "ofs" isn't pointer.
> I can't understand this "polymorphic" action.
> Maybe type conversion happend?

An ofstream reference is freely convertible to an ostream reference.
Since there is no operator<< that accepts an ofstream (as value or
reference and a CPerson (as value or reference,) the compiler uses the
operator<< that you provided.

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shnya <testmhigashi> wrote:
> I don't know why following code is working.
>
> #include <iostream>
> #include <fstream>
>
> using namespace std;
>
> istream&
> operator<<(ostream& os, CPerson &str){
> return os << CPerson.getName();
> /* getName returns const char* */
> }

This should probably return ostream& instead of istream&, in order to
allow chaining of output commands like:

cout << person1 << '\n' << person2 << '\n';

> CPerson person; //Person is User-Definition Class
>
> // not strange
> cout << person;
>
> ofstream ofs("test.txt");
>
> // strange. why do this code work?
> ofs << person;
>
[snip]
>
> ofstream is inherited ostream.

Yes, ofstream is a derived class of ostream.

> but "operator<<" is not member operator, no virtual directive, and
> ofstream object "ofs" isn't pointer.
> I can't understand this "polymorphic" action.
> Maybe type conversion happend?

In C++, the line:

ofs << person;

is shorthand for:

operator<<(ofs, person);

Since you are passing the ostream& to operator<< by reference (that is
what the '&' means in the function declaration), then that means that
*any* class derived from ostream is allowed to be passed to operator<<.
Passing by reference is almost like passing by pointer, except that the
pointer is always dereferenced, and you cannot change the object that
the reference refers to.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

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In article <1181250320.122666.49440@g37g2000prf.googlegroups.com>,
shnya <testmhigashi> wrote:

>
> using namespace std;
>
> istream&
> operator<<(ostream& os, CPerson &str){
> return os << CPerson.getName();
> /* getName returns const char* */
> }
assuming this operator << is really
ostream & operator << (ostream &os,const CPerson &str)
{
return os << str,getName();
}

all should work

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On Jun 8, 12:58 pm, Carl Barron <cbarron...@adelphia.net> wrote:
> In article <1181250320.122666.49...@g37g2000prf.googlegroups.com>,
>
> shnya <testmhiga...> wrote:
>
> > using namespace std;
>
> > istream&
> > operator<<(ostream& os, CPerson &str){
> > return os << CPerson.getName();
> > /* getName returns const char* */
> > }
>
> assuming this operator << is really
> ostream & operator << (ostream &os,const CPerson &str)
> {
> return os << str,getName();
>
> }
>
> all should work
>

Dear Daniel T. and Dear Carl Barron.

I'm sorry.
This is my mistake.

Yes, your code is what I want to write.
I will take care when I post next.
Thank you.

Dear Marcus Kwok.

Thank you.
I want to hear this reason.
I thought that I should need following code.

ofstream &
operator<<(ofstream& os, CPerson &str){
return os << CPerson.getName();
}

Because I thought that reference was similar to the value type, but
reference wasn't copied and point same object.
So I check following code.

----
#include <iostream>

using namespace std;

class Base {
string name_a;
public:
explicit Base(string i) :name_a(i) {}
virtual string getName(){ return name_a;}
};

class Derived : public Base {
string name_b;
public:
explicit Derived(string i, string j) :name_b(i), Base(j) {}
virtual string getName(){ return name_b;}
};

void
printValue(Base a){
cout << a.getName() << endl;
}

void
printRef(Base &a){
cout << a.getName() << endl;
}

void
printPtr(Base *a){
cout << a->getName() << endl;
}

int
main(void){

Base a("class a");
Derived b("class b", "class a");

printValue(a);
printValue(b);
printRef(a);
printRef(b);
printPtr(&a);
printPtr(&b);

return 0;
}

----
output

class a
class a
class a
class b
class a
class b

I seem that I don't understand elements of C++.
Lastly, I thank you all again for answering my question.


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