Question

As in this question, I’m experimenting to stream via a class using SBRM/RAII, so

SBRM(x) << "test";

could do some extra’s in the destructor, but my template knowledge seems to be limited.

What I have (made simpler for clarity) is:

#include <iostream>
#include <sstream>

class SBRM
{
public:
  SBRM(int j) : i(j) {}
  ~SBRM() { std::cout << "SBRM(" << i << "): " << oss.str() << std::endl; }

  template<typename T> SBRM& operator<<(T& in) { oss << in; return *this; }
  // SBRM& operator<<(const long long& in) { oss << std::hex << "0x" << in; return *this; }
  SBRM& operator<<(const double& in) { oss << in; return *this; }
  SBRM& operator<<(const void* in) { oss << in; return *this; }

private:
  int i;
  std::ostringstream oss;
};


int main()
{
  std::string ttt = "world";
  const int i = 3;
  SBRM(1) << "Hello";
  SBRM(2) << ttt;
  SBRM(3) << 0x1234567890123ll; 
  SBRM(4) << &i;
  SBRM(5) << 5;
  SBRM(6) << 0.23;
  SBRM(7) << i;
  SBRM(8) << 5 << ", " << ttt << ", " << &i;
}

This sort of works:

SBRM(1): Hello
SBRM(2): world
SBRM(3): 3.20256e+14
SBRM(4): 0xbf8ee444
SBRM(5): 5
SBRM(6): 0.23
SBRM(7): 3
SBRM(8): 5, world, 0xbf8ee444

but my main concern is: why does the compiler require me to overload the template when using (non-string) literals?
Are there any tricks to avoid this or am I taking a wrong approach?
Other suggestions are welcome because I now resorted to using a macro for

NOT_QUITE_SBRM_MACRO(3, "At least, " << 5 << ", this works");

The issue is seen with gcc 4.1.2. and 4.4.3. Without the overloaded functions, I get:

sbrm-stream.cpp: In function ‘int main()’:
sbrm-stream.cpp:27: error: no match for ‘operator<<’ in ‘SBRM(3) << 320255973458211ll’
sbrm-stream.cpp:10: note: candidates are: SBRM& SBRM::operator<<(T&) [with T = long long int]
sbrm-stream.cpp:28: error: no match for ‘operator<<’ in ‘SBRM(4) << & i’
sbrm-stream.cpp:10: note: candidates are: SBRM& SBRM::operator<<(T&) [with T = const int*]
...

Answer 1

Because you’re expecting a non-const argument and literals can never be treated as such. Make the argument const and your troubles will go away:

template<typename T> SBRM& operator<<(T const& in) { oss << in; return *this; }

And as David has mentioned in his comment, you need overloads when using manipulators such as endl. Here’s a shot at them:

SBRM& operator <<(std::ostream& (*manip)(std::ostream&)) {
    oss << manip; // alternatively: manip(os);
    return *this;
}

// same for:

ostream& operator <<(ios& (*manip)(ios&));
ostream& operator <<(ios_base& (*manip)(ios_base&));

This covers all the parameterless manipulators.

I’m not actually sure how the parametrized manipulators from <iomanip> work but they seem to return a proxy object that can use the generic operator << variant.