Re: endl like behavior

From:
"mlimber" <mlimber@gmail.com>
Newsgroups:
comp.lang.c++
Date:
13 Jun 2006 08:08:32 -0700
Message-ID:
<1150211312.835523.12630@h76g2000cwa.googlegroups.com>
schwehr@gmail.com wrote:

I also tried creating a function like this...

namespace std {
slog& endl(slog& s) {
  std::cerr << "endl called" << std::endl;
  return s;
}
} // std

The compiler then complains with this which says that I am clearly
missing something :)

make && ./slogcxx-test
g++ -o slogcxx-test slogcxx.cpp -Wall -g -DTESTING
slogcxx.cpp: In function 'int main(int, char**)':
slogcxx.cpp:36: error: no match for 'operator<<' in 'log << std::endl'
slogcxx.cpp:23: note: candidates are: slog& operator<<(slog&, const
int&)
make: *** [slogcxx-test] Error 1


First, please quote the message you are replying to. It helps everyone
to follow the conversation.

Second, you may not add things to namespace std.

Third, what you're missing is an operator that will use the iomanip.
Here's what you're looking for:

 #include <iostream>
 using namespace std;

 class slog {}; // end slog class

 slog& operator<< (slog &s, const int r)
 {
   cerr << r;
   return s;
 }

 // Here's the key:
 slog& operator<< ( slog &s, slog&(*iomanip)(slog&) )
 {
   iomanip( s );
   return s;
 }

 slog& endl( slog &s )
 {
   cerr << endl;
   return s;
 }

 int main()
 {
   slog log;
   log << 1 << endl;
 }

Cheers! --M

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