Newbie: How to store data into my object with << ?

[Goran]

Hi all,

I want to store data with << like this way:

myClass_t aObject;
string aString("foo");

// no output, just edit and store the string
aObject << aString;

// here comes the edited string
cout << aObject;

My question is:
How do I realize "aObject << aString;"? I've read a lot about
overloading operators and it's no prob. But what's the signature of
the operator which realizes "aObject << aString;"?

Thanks a lot

Goran

[gnuyuva]

On May 15, 5:56 pm, Goran <[e-mail]:

Hi all,

I want to store data with << like this way:

myClass_t aObject;
string aString("foo");

// no output, just edit and store the string
aObject << aString;

// here comes the edited string
cout << aObject;

My question is:
How do I realize "aObject << aString;"? I've read a lot about
overloading operators and it's no prob. But what's the signature of
the operator which realizes "aObject << aString;"?

MyClass_t& MyClass_t::operator<<(const std::string& string)
{ ... }

[Victor Bazarov]

gnuyuva wrote:

[..]
MyClass_t& MyClass_t::operator<<(const std::string& string)
{ ... }

{ ...
return *this; // idiomatic
}

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Jim Langston]

Victor Bazarov wrote:

gnuyuva wrote:

>[..]
MyClass_t& MyClass_t::operator<<(const std::string& string)
{ ... }

{ ...
return *this; // idiomatic
}

V

A question on this. I checked the own signature in my own code and found
this:

void CItem::operator<<(const std::string& sIn)
{
// ...
}

which, as far as my usage of it, is working fine. Although it's possible
that this is some extention, I've only compiled it under one compiler.

Is my signature wrong, another way to do it, poor design, okay, or?

My usage being like this:

std::istream& operator>>( std::istream& is, CItem& Item )
{
std::string Line;
if ( getline( is, Line ) )
Item << Line;
return is;
}

What purpose would/could be served by returning a referece to the instance?
Maybe returning it from some function?

return Item << Line;

?

--
Jim Langston
ta*******@rocketmail.com

[Jerry Coffin]

In article <2ac1ff02-7630-4bf9-b60d-
43**********@l42g2000hsc.googlegroups.com>, po**********@gmail.com
says...

Hi all,

I want to store data with << like this way:

myClass_t aObject;
string aString("foo");

// no output, just edit and store the string
aObject << aString;

// here comes the edited string
cout << aObject;

My question is:
How do I realize "aObject << aString;"? I've read a lot about
overloading operators and it's no prob. But what's the signature of
the operator which realizes "aObject << aString;"?

You've already gotten one answer, but I'll add another that overloads
operator<< with a non-member function:

myClass_t &operator<<(myClass_t &sink, aString const &source);

As for why you return an instance of the stream-like object, it's to
allow chaining the operator, so you can do things like:

myCLass_t aObject;
string aString("foo"), bString("bar");

aObject << aString << bString;

--
Later,
Jerry.

The universe is a figment of its own imagination.

[Rolf Magnus]

Jim Langston wrote:

Victor Bazarov wrote:

>gnuyuva wrote:

>>[..]
MyClass_t& MyClass_t::operator<<(const std::string& string)
{ ... }

{ ...
return *this; // idiomatic
}

V

A question on this. I checked the own signature in my own code and found
this:

void CItem::operator<<(const std::string& sIn)
{
// ...
}

which, as far as my usage of it, is working fine.

Why shouldn'it?

Although it's possible that this is some extention,

What?

I've only compiled it under one compiler.

Is my signature wrong, another way to do it, poor design, okay, or?

Oh, you mean because you returned void? No, that's fine.

My usage being like this:

std::istream& operator>>( std::istream& is, CItem& Item )
{
std::string Line;
if ( getline( is, Line ) )
Item << Line;
return is;
}

What purpose would/could be served by returning a referece to the
instance? Maybe returning it from some function?

return Item << Line;

?

It makes it possible to concatenate several operator calls:

Item << Line << Line2;

[Victor Bazarov]

Jerry Coffin wrote:

[..] I'll add another that overloads
operator<< with a non-member function:

myClass_t &operator<<(myClass_t &sink, aString const &source);

Usually operators that change the object should probably be members of
the class (IMNSHO); the whole point of making some operators non-members
is to allow conversions applied to the left-hand-side operand (which is
not the case here, I guess).

[..]

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Jerry Coffin]

In article <g0**********@news.datemas.de>, v.********@comAcast.net
says...

Jerry Coffin wrote:

[..] I'll add another that overloads
operator<< with a non-member function:

myClass_t &operator<<(myClass_t &sink, aString const &source);

Usually operators that change the object should probably be members of
the class (IMNSHO); the whole point of making some operators non-members
is to allow conversions applied to the left-hand-side operand (which is
not the case here, I guess).

One point is to allow conversions -- another is to allow extension
without modification.

--
Later,
Jerry.

The universe is a figment of its own imagination.

[Jim Langston]

Rolf Magnus wrote:

Jim Langston wrote:

>Victor Bazarov wrote:

>>gnuyuva wrote:
[..]
MyClass_t& MyClass_t::operator<<(const std::string& string)
{ ... }

{ ...
return *this; // idiomatic
}

V

A question on this. I checked the own signature in my own code and
found this:

void CItem::operator<<(const std::string& sIn)
{
// ...
}

which, as far as my usage of it, is working fine.

Why shouldn'it?

>Although it's possible that this is some extention,

What?

>I've only compiled it under one compiler.

>Is my signature wrong, another way to do it, poor design, okay, or?

Oh, you mean because you returned void? No, that's fine.

>My usage being like this:

std::istream& operator>>( std::istream& is, CItem& Item )
{
std::string Line;
if ( getline( is, Line ) )
Item << Line;
return is;
}

What purpose would/could be served by returning a referece to the
instance? Maybe returning it from some function?

return Item << Line;

?

It makes it possible to concatenate several operator calls:

Item << Line << Line2;

Ahh, yes. That clears it up then. In my case that line is (and should be)
an error. So void is perfectly fine for me. Thanks.

--
Jim Langston
ta*******@rocketmail.com