What this means?

[PencoOdStip]

void print_msg( ostream &os, const string &msg )
{
if ( msg.empty() )
// nothing to print; terminate function ...
return;

os << msg;
}

I don't understand the first line : void print_msg( ostream &os, const
string &msg )

by writing ostream &os,you do what ? what that means? same goes for
const string &msg ...

why not just const string msg,without the & ?

 

[Victor Bazarov]

void print_msg( ostream &os, const string &msg )
{
if ( msg.empty() )
// nothing to print; terminate function ...
return;

os << msg;
}

I don't understand the first line : void print_msg( ostream &os, const
string &msg )

by writing ostream &os,you do what ? what that means? same goes for
const string &msg ...

why not just const string msg,without the & ?

What book on C++ are you reading that doesn't explain references?

The declaration

T &t;

means that 't' is a reference to an object of type 'T'. Look them up.

V

 

[Devon Null]

void print_msg( ostream &os, const string &msg )
{
if ( msg.empty() )
// nothing to print; terminate function ...
return;

os << msg;
}

I don't understand the first line : void print_msg( ostream &os, const
string &msg )

by writing ostream &os,you do what ? what that means? same goes for
const string &msg ...

why not just const string msg,without the & ?

It passes the address of the parameters the function is called with. It
allows you to directly alter the data with out having to return a
specified value. It is much easier to handle that way when doing
complicated action, if you are careful. Also it is much more efficient -
less overhead.

 

[Gavin Deane]

It passes the address of the parameters the function is called with.

No it doesn't. That answer would be appropriate if the OP had asked
about pointers, but they didn't - they asked about references.

Gavin Deane

 

[Salt_Peter]

void print_msg( ostream &os, const string &msg )
{
if ( msg.empty() )
// nothing to print; terminate function ...
return;

os << msg;

}

I don't understand the first line : void print_msg( ostream &os, const
string &msg )

by writing ostream &os,you do what ? what that means? same goes for
const string &msg ...

why not just const string msg,without the & ?

Because 'const string msg' would create a copy of the string being
passed. Therefore invoking a costly copy constructor. Instead of
constructing the string; pass it by const reference: 'const string&
msg'. This creates an 'alias' of the original which can't be modified
nor reseated.

In C++, unless there is reason to do otherwise, passing by reference
should be the default.

This is specially critical if you consider 'ostream& os' in your
function. You would definitely not enjoy the results of copy
constructing a standard output stream (std:stream) since you are
probably passing your precious std::cout to that function to display
messages on the console.

I'ld suggest writing:

void print_msg( ostream& os, const string& msg ) { ... }

.... instead of the confusing ...

void print_msg( ostream &os, const string &msg ) { ... }

 

[PencoOdStip]

So what is the difference between these two?

void print_msg( ostream& os, const string& msg ) { ... }

void print_msg( ostream &os, const string &msg ) { ... }

 

[Dave Steffen]

So what is the difference between these two?

void print_msg( ostream& os, const string& msg ) { ... }

void print_msg( ostream &os, const string &msg ) { ... }

Semantically, none whatsoever. The top version is usually, but not
always, preferred by C++ programmers. The bottom version is
usually, but not always, preferred by C programmers.

What book are you reading that doesn't explain this? Get a better
one!


--
Dave Steffen, Ph.D. A Zen master once said to me, "Do the
Software Engineer IV opposite of whatever I tell you."
Numerica Corporation So I didn't.
ph (970) 461-2000 x227
-- not Hofstadter (but should have been)