Re: delayed eof()?

From:

"Daniel T." <postmaster@verizon.net>

Newsgroups:

comp.lang.c++

Date:

Fri, 12 May 2006 16:20:46 GMT

Message-ID:

<postmaster-8E7AE2.12204512052006@news.west.earthlink.net>

In article <e41rni$3t7$1@hahn.informatik.hu-berlin.de>,
Thomas Urban <thomas.urban@toxa.de> wrote:

Hi,

I'm probably asking something posted several times before and maybe you
feel annoyed by answering ... please excuse me nevertheless, as I need
an urgent statement on whether and why this case is conforming to "STL
standard" (read as "is conforming to intentions of STL designers") or not:

Taking this minimal example

------
#include <iostream>
#include <sstream>

using namespace std;

int main( void )
{
    stringstream s;
    char ch;

    s.str( "Hello" );

    while ( !s.eof() )
    {
        s >> ch;
        cout << ch;
    }

    cout << endl;

}
------

results in single line output

Helloo

(sending 6 instead of 5 characters to cout). I tried this with MS Visual
C++ 2005 Express Edition and GNU 4.0.x with libstdc++ 4.0.x ... they
behave equivalently.

What's so wrong about using this simple algorithm to read from string
char-wise, e.g. for parsing it ...

The code is behaving exactly as expected. Follow the logic for a second.
Let's use s.str( "1" ) though to make things faster...

eof() is false so enter loop:
s >> ch; reads '1' from the loop
cout << ch; prints '1'
eof() is still false (we have not reached the end of the stream yet.
s >> ch; reads 'eof' *now* eof() is true.
cout << ch prints '1' again.

Try this instead:

while ( s >> ch ) {
cout << ch;
}