cartec69@gmail.com wrote:
On Thursday, June 21, 2012 7:59:23 AM UTC-5, Gerhard Fiedler wrote:
I'm still working on getting my original approach to work (please see
my other post in this thread if you're interested). I'm not yet
understanding well enough how I can limit the matching.
See http://stackoverflow.com/questions/257288/is-it-possible-to-write-a-c-template-to-check-for-a-functions-existence
for a very good discussion of how to detect a member function.
Applying the principles discussed there to your program:
#include<iostream>
#include<type_traits>
template<typename T>
class is_printable
{
typedef char yes[1];
typedef char no[2];
template<typename C> static yes& test( decltype(&C::print) );
template<typename C> static no& test(...);
public:
static const bool value = (sizeof(test<T>(0)) == sizeof(yes));
};
template< class Printable>
typename std::enable_if<is_printable<Printable>::value,std::ostream&>::type
operator<<( std::ostream&stream, Printable const&printable )
{
printable.print( stream );
return stream;
}
class MyClass
{
public:
MyClass( char const *val ) : str( val ) {}
void print( std::ostream&stream ) const { stream<< str; }
private:
char const *str;
};
int main(int argc, char* argv[])
{
std::cout<< MyClass( "bar" )<< std::endl;
}
If you have types in your codebase that you want to exclude - say they
have a print member that serves some other purpose - you can do so
with a specialization:
template<>
class is_printable<Foo>
{ public: static const bool value = false; };
Thanks!
At first I thought this works as I want it, but then I ran into this
snag. Google Test defines a function like this:
template<typename T>
void GTestStreamToHelper(std::ostream* os, const T& val) {
*os<< val;
}
I invoke it like this (for testing resolution):
int main(int argc, char* argv[])
{
MyClass foo( "bar" );
std::cout<< "test "<< foo<< std::endl;
std::string const gtest( "gtest" );
GTestStreamToHelper(&std::cout, gtest );
std::cout<< std::endl;
}
For some reason, my operator<<() gets chosen for the '<<' inside
GTestStreamToHelper, then it complains that std::string doesn't have a
print() method. Shouldn't it use a different operator<<() definition?
It will use the template that matches better. Unless the compiler has a
bug in it, that is. Please post the shortest code to replicate the
situation. Do so every time you need help with code not compiling or