Re: std::istreambuf_iterator

From:
 James Kanze <james.kanze@gmail.com>
Newsgroups:
comp.lang.c++
Date:
Sun, 12 Aug 2007 14:46:16 -0000
Message-ID:
<1186929976.603257.213370@d55g2000hsg.googlegroups.com>
On Aug 11, 11:51 pm, "BobR" <removeBadB...@worldnet.att.net> wrote:

James Kanze <james.ka...@gmail.com> wrote in message...

On Aug 11, 6:16 pm, "BobR" <removeBadB...@worldnet.att.net> wrote:

James Kanze <james.ka...@gmail.com> wrote in message...

    std::vector< std::string > data(
            (std::istreambuf_iterator< std::string >( inputFile )),
            (std::istreambuf_iterator< std::string >()) ) ;
(Formally, it's sufficient to add the parentheses to just one of
the arguments, but I like orthogonality.)

Did that compile for you ( using std::string in
istreambuf_iterator!! )? (which compiler?)


You're right, I didn't try it:-). You'd need
std::istream_iterator for std::string.

 Don't know what I was thinking of.


Probably that thread a while back. But, that was using type 'char'.


Not really. I was basing my response on the original posting,
plus the error message he got. The error message was due to his
declaring a function, not a variable. (And of course, declaring
std::istreambuf_iterator< std::string > as a parameter of a
function probably won't cause any errors from the compiler,
since using a class template as a function parameter in a
declaration doesn't trigger instantiation.)

// ref: std::ifstream PicIn( ...., '::in | '::binary );

     std::vector<unsigned char> Image(
         std::istreambuf_iterator<char>( PicIn.rdbuf() ),
         std::istreambuf_iterator<char>()
         );

The code you showed was the *first* thing I tried. <G>


Which, of course, would generate a different error message than
the one he got. So he really had two problems. The data
definition which is really a function declaration is so common
that I didn't think further, but that's not really an excuse for
posting an answer without having actually tried it. (Formally,
std::istreambuf_iterator< std::string > is undefined behavior,
so you're not guaranteed a compiler error. But I'd be very
surprised if you didn't get one if you actually tried to use it.

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