Re: istream && istream_iterator behavior

[Phil Edwards <pedwards at disaster dot jaj dot com>]

I swear I sent this mail already...  okay maybe I'm insane.


On Fri, Feb 23, 2001 at 12:58:25PM +0100, Laurent Marzullo wrote:
> $ g++ essai.cpp
> $ ./a.out
> 1234
> read1 = `1'
> read2 = `2'	// ?????? I was expected `1' too
> 
> May someone explain me why I've got this output and not
> read2 = `1' ??
> 
> Is this a correct behavior ?

I believe this is correct.  Here's what's happening:

When an istream_iterator<char>(s) is constructed, one character is read
and cached from s.  This means that s moves forward to the next character.

You have read1's l_in_ptr being constructed, which reads and caches '1'.
Then read2's l_in_ptr is constructed, which reads and caches '2'.  If read1
were to use

    cout << "read1 = `" << (*l_in_ptr) << "'\n";
    cout << "read1 = `" << (*l_in_ptr) << "'\n";

twice before calling read2(), then I would expect '1' to be printed both
times, because the cached character is being printed both times.

The Standard /allows/ the cached character to be read at construction time,
but does not /require/ it, so do not rely on this behavior.


Luck++;
Phil

-- 
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