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Why is the compiler complain in stringstream?

Posted on 2012-04-13
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Last Modified: 2012-04-17
the read() function in the stringstreram class returns a istream& type. So if I do a call like this...

std::istream readStream =  mDataStream.read(buffer, buffersize);

I am getting a compilation error "synthesized method ‘std::basic_istream<char, std::char_traits<char> >::basic_istream(const std::basic_istream<char, std::char_traits<char> >&)"

Then I change to like this.

std::istream* readStream =  mDataStream.read(buffer, buffersize);

Then I get this error....
invalid conversion from ‘void*’ to ‘std::istream*

What I am doing wrong?
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Question by:prain
9 Comments
 
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Expert Comment

by:aikimark
Comment Utility
what language and development tool are you using?
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Expert Comment

by:cup
Comment Utility
Take this program, change it to the way you're using it and post it back.  It is difficult to visualize what your problem is.  This one works in Visual studio

#include <sstream>

struct Dummy
{
   std::istringstream mResult;
   std::istream* read (char* buffer, int buffersize)
   {
      return &mResult;
   }
} mDataStream;

int main ()
{
   char buff[1024];
   std::istream* readStream = mDataStream.read (buff, sizeof(buff));
   return 0;
}

Open in new window

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Author Comment

by:prain
Comment Utility
I am developing on Unix/g++. I am confused with what the compiler says.
If I use the function without the return type, it's great.
For example...

mDataStrea.read(buffer, bufferSize);

But if I capture the return type, I get this problem...
std::istream* readStream =  mDataStream.read(buffer, buffersize);

error: invalid conversion from ‘void*’ to ‘std::istream*’

BTW, my mDataStream is a C++ stringstream instance. So unfortunately I cannot use your code in my C++ class. I have to stick to the stringstream instance.
-prain
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Author Comment

by:prain
Comment Utility
Someone try this simple example and tell me why the compiler is barking..
BTW, I using Unix/g++

#include <string>
#include <iostream>
#include <sstream>

int main ()
{
   std::stringstream sstr;

   sstr << "Hello" << " World" ;

   char buffer[100];
   int size = 11;

   std::istream* aIStream = sstr.read(buffer, size);


  return 0;
}
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Author Comment

by:prain
Comment Utility
BTW, since stringstream inherits istream, the read() in istream is defined as this:

istream& read ( char* s, streamsize n );
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Accepted Solution

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Infinity08 earned 125 total points
Comment Utility
>> std::istream readStream =  mDataStream.read(buffer, buffersize);

Copying streams is not defined.


>> std::istream* readStream =  mDataStream.read(buffer, buffersize);

You can't assign a reference to a pointer.


Try this :

std::istream& readStream =  mDataStream.read(buffer, buffersize);

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Author Closing Comment

by:prain
Comment Utility
Thanks
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Expert Comment

by:sarabande
Comment Utility
FYI:

it is intended by design that a stream cannot be copied. they explicitly didn't provide a copy constructor (and no assignment operator) cause a stream is not a normal container but has status and position members stored for streaming purposes. in your case the read member function returns a reference to istream while the stream you used was a stringstream. so you don't get a reference to the original stream but only to the istream baseclass.

the reasons why the read member function returns a reference to istream is not mainly to assign the return value to a second variable (you still could/should use the first variable) but for two other purposes. one is that the read would return NULL in case the read operation failed (error or end-of-stream), what makes it possible to do a read like

if (!mDataStream.read(buffer, buffersize))
{
    ...

Open in new window


or to use a read in a while statement.

the other is that you could use the returned stream reference to pass it to next stream operation:

(mDataStream.read(buffer, 4)).read(&buffer[4], 2);

Open in new window


what in case of read looks strange but was often used for operator>> like in

if (mDataStream >> i >> j)

Open in new window


to do a formatted read into more variables with one statement.

Sara
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