Question about ostream

[=?ISO-8859-1?Q?Ney_Andr=E9_de_Mello_Zunino?=]

You're trying to _copy_ a stream.

That doesn't make sense.

Because it doesn't make sense the std:stream class disables copying (no
copy constructor), and you'll have to pass that stream by reference.

Indeed and I would expect that compiler's diagnostic to be more clear
about it. The error message given is quite misleading, if not nonsensical.

 

[Robert Sherry]

It is my understanding that the standard variable cout is of type
ostream. Please consider the following C++ program.

#include <iostream>

using namespace std;

void func1( ostream os1 );

main()
{
func1( cout );
}

void
func1( ostream os1 )
{
}

When I compile this program using the Microsoft Visual C++ Studio, the
compiler generates an error on the call to func1. The message is:
c:\dev\ostreamTest\main.cpp(9): error C2664: 'func1' : cannot
convert parameter 1 from 'std:stream' to 'std:stream'

I find it hard to believe that I have found a bug in the compiler but I do
not know what is wrong. Since the types of the formal and actual parameters
are the same, why is any conversion needed? I am hoping for some help from
the group.

Robert Sherry

 

[Alf P. Steinbach]

* Robert Sherry:

It is my understanding that the standard variable cout is of type
ostream. Please consider the following C++ program.

#include <iostream>

using namespace std;

void func1( ostream os1 );

main()
{
func1( cout );
}

void
func1( ostream os1 )
{
}

When I compile this program using the Microsoft Visual C++ Studio, the
compiler generates an error on the call to func1. The message is:
c:\dev\ostreamTest\main.cpp(9): error C2664: 'func1' : cannot
convert parameter 1 from 'std:stream' to 'std:stream'

I find it hard to believe that I have found a bug in the compiler but I do
not know what is wrong. Since the types of the formal and actual parameters
are the same, why is any conversion needed? I am hoping for some help from
the group.

You're trying to _copy_ a stream.

That doesn't make sense.

Because it doesn't make sense the std:stream class disables copying (no
copy constructor), and you'll have to pass that stream by reference.

 

[Sumit Rajan]

It is my understanding that the standard variable cout is of type
ostream. Please consider the following C++ program.

#include <iostream>

using namespace std;

void func1( ostream os1 );

void func1(ostream& os1);

main()

int main()

{
func1( cout );
}

void
func1( ostream os1 )

void func1(ostream& os1)

{
}

Regards,
Sumit.

 

[Robert Sherry]

I wish to thank the group for their fast response. It was very helpful.

Robert Sherry

 

[Artie Gold]

* Robert Sherry:



You're trying to _copy_ a stream.

That doesn't make sense.

Because it doesn't make sense the std:stream class disables copying (no
copy constructor), and you'll have to pass that stream by reference.

Just to make things a little more clear, make that:

void func1( ostream& os1 );

and

void func1( ostream& os1 ) {}

The terminology we usually use, i.e. `pass it by reference' must be
terribly confusing for C++ newbies; passing by value and passing by
reference are indistinguishable at the function call site. Perhaps we
should start saying that the variable should be `received by reference'
instead?

Just a thought.

Cheers,
--ag

 

[lixiaoyao]

when I compile this code,it generate the following rsult,I am a little
bit wondering.
[library]$ g++ ofstream.cpp&
"[4] 2864"
why generate this?
thanks

 

[Alf P. Steinbach]

* lixiaoyao:

when I compile this code,it generate the following rsult,I am a little
bit wondering.
[library]$ g++ ofstream.cpp&
"[4] 2864"
why generate this?

The Unix shell is off-topic in this group.