Converting a struct to a stream

[Jacek Dziedzic]

I have created the following conversion in the header file
for my struct called SomeStruct.

operator std::string () {return "some useless text";}

I'm using it as follows.

SomeStruct someS;
std:cout << (std::string)someS;

This far, everything works just fine. Now, i got lazy of
constantly writing the string-part so i thought i'd like to
extend my struct to be convertible to a stream. So i
went as follows.

operator std:stream () {return "some other text";}

This didn't work, however. I don't understand the error
messages either. Any hints?

Also, would it be a smarter idea to overload << operator
for my struct?

Definitely.

How do i do that? I've played around
with friend-keyword but frankly, i must be missing
something, getting nowhere. Any help is appreciated...

You need a standalone function, not a member of a
class. Something like

ostream& operator<<(ostream& whither, const yourstruct& what) {
whither << "some useless text";
return whither;
}

HTH,
- J.

 

[The Cool Giraffe]

I have created the following conversion in the header file
for my struct called SomeStruct.

operator std::string () {return "some useless text";}

I'm using it as follows.

SomeStruct someS;
std:cout << (std::string)someS;

This far, everything works just fine. Now, i got lazy of
constantly writing the string-part so i thought i'd like to
extend my struct to be convertible to a stream. So i
went as follows.

operator std:stream () {return "some other text";}

This didn't work, however. I don't understand the error
messages either. Any hints?

Also, would it be a smarter idea to overload << operator
for my struct? How do i do that? I've played around
with friend-keyword but frankly, i must be missing
something, getting nowhere. Any help is appreciated...

 

[Fei Liu]

I have created the following conversion in the header file
for my struct called SomeStruct.

operator std::string () {return "some useless text";}

I'm using it as follows.

SomeStruct someS;
std:cout << (std::string)someS;

This far, everything works just fine. Now, i got lazy of
constantly writing the string-part so i thought i'd like to
extend my struct to be convertible to a stream. So i
went as follows.

operator std:stream () {return "some other text";}

This didn't work, however. I don't understand the error
messages either. Any hints?

Also, would it be a smarter idea to overload << operator
for my struct? How do i do that? I've played around
with friend-keyword but frankly, i must be missing
something, getting nowhere. Any help is appreciated...

Try operator std:stream & (std:stream & os) {return os << "some
other text";}

First of all you can't construct std:stream from char *, and secondly
std:stream is not copy constructible. Your best bet is to override
'<<' such as:
operator std:stream & << (std:stream & os) { return os << "some text"; }

Fei

 

[Ralph D. Ungermann]

You need a standalone function, not a member of a
class. Something like

ostream& operator<<(ostream& whither, const yourstruct& what) {

And probably (if you need access to private members):

class yourstruct
{
// your stuff
friend ostream& operator<<(ostream&, const yourstruct&);
}

 

[The Cool Giraffe]

Ralph D. Ungermann wrote/skrev/kaita/popisal/schreibt :

Jacek Dziedzic wrote:

And probably (if you need access to private members):

class yourstruct {
// your stuff
friend ostream& operator<<(ostream&, const yourstruct&); }

Thanks. Now, some wonderings i have about your answer.
1. What happened to the names of the parameters? I was
expecting something like the below.

class yourstruct {
// your stuff
friend ostream& operator<<(ostream& param1, const yourstruct& param2); }

2. How to implement the method itself? I'm guessing
somewhere in the lines of this:

param1 = "something";
return param1;

or perhaps

param1 << "something";
return param1;

but is the above a legal C++ code? I know some things
do work but are not really that good to use. Hence my
wondering. How would YOU implement that function?

 

[Jacek Dziedzic]

Ralph D. Ungermann wrote/skrev/kaita/popisal/schreibt :


Thanks. Now, some wonderings i have about your answer.
1. What happened to the names of the parameters? I was
expecting something like the below.

In declarations you can omit them.

2. How to implement the method itself? I'm guessing
somewhere in the lines of this:

param1 = "something";
return param1;

I don't think you can assign a string to a stream.

or perhaps

param1 << "something";
return param1;

This one.

but is the above a legal C++ code?

This one is. Remember to return the stream, this
will allow for chaining multiple <<'s together.

> I know some things
do work but are not really that good to use. Hence my
wondering. How would YOU implement that function?

Just like you did in:

> param1 << "something";
> return param1;

BTW, you are not "converting" a struct to a stream,
rather sending it there or perhaps serializing would be
a better word.

HTH,
- J.

 

[Ralph D. Ungermann]

Ralph D. Ungermann wrote/skrev/kaita/popisal/schreibt :

Jacek Dziedzic wrote:

You need [...] Something like

ostream& operator<<(ostream& whither, const yourstruct& what) {

And probably (if you need access to private members):

class yourstruct {
// your stuff
friend ostream& operator<<(ostream&, const yourstruct&); }

Thanks. Now, some wonderings i have about your answer.
1. What happened to the names of the parameters?

Feel free to provide argument names as you like. They have no effect,
unless you _define_ the function; in the body, these are the names you
have to use in your implementation (*).

I've omitted the names, because the types are self-explaining here.
But I'd definitely provide names there:
bool resize( int, int, int, int, bool );

2. How to implement the method itself?

ostream& operator<<(ostream& os, const yourstruct& what)
{
os << "{";
if ( what.ok() )
os << what.private_x << ", " << what.private_y;
else
os << "(undefined)";
os << "}"; // definitely NO std::endl here!

return os;
}

N.B.: Stream modifiers probably won't work as expected:
std::cout << std::setw(8) << yourstruct(17, 4) << std::endl;
// will look ugly

 

[The Cool Giraffe]

Jacek Dziedzic wrote/skrev/kaita/popisal/schreibt :

In declarations you can omit them.


I don't think you can assign a string to a stream.


This one.


This one is. Remember to return the stream, this
will allow for chaining multiple <<'s together.


Just like you did in:


BTW, you are not "converting" a struct to a stream,
rather sending it there or perhaps serializing would be
a better word.

Right, got it. Thank you very much.