Re: Template functions and avoiding implicit conversions
* elazro:
On Feb 28, 5:49 am, "James Kanze" <james.ka...@gmail.com> wrote:
On Feb 28, 10:22 am, "Alf P. Steinbach" <a...@start.no> wrote:
* tarjei.knaps...@gmail.com:
Assuming the above mentioned corrections to the class definition, when
you write what you wrote here you /do not/ get a an implicit conversion
to bool.
#include <iostream>
#include <ostream>
#include <string>
void say( char const s[] ) { std::cout << s << std::endl; }
class CommandLineOptions
{
public:
template <typename ValueType>
void Add( std::string& arg, bool required )
{ say( "bool" ); }
template <typename ValueType>
void Add( std::string const& arg, ValueType const& aDefault )
{ say( "T" ); }
};
int main()
{
CommandLineOptions().Add<std::string>( "method", "invert" );
}
Output: "T".
Now I'm lost. Both my compilers agree with you, so I'm
obviously missing something, somewhere, but I don't understand.
I even tried it with non-template functions (replacing ValueType
with std::string), and got the same results. But why?
I don't think the above code snippet is well formed - the first
overload has a non-const std::string reference as its first argument -
the same problem as the original coded.
Rather embarassing (even if Andy Little thought it intentional, giving
me credit for smarts instead of dumbs) -- I'm not sure how that happened.
So, now it seems I'm obliged to solve this thing, that is, without
complicated SFINAE, or even without Carl Barron's elegant trick
(requires library or extra definitions).
The main problem is underspecification: should
clo.Add<std::string>( "method", true );
be supported or not (i.e. what's the template parameter actually used for)?
Tarjei didn't say, but his example had the second argument's type be the
template parameter type, and bool isn't std::string, so, probably not.
I'll assume not.
Thus the second overload will
always be a better match (well, the only match) when the first
argument is a const char *. If you replace main with:
int main()
{
CommandLineOptions clo;
clo.Add<std::string>("method",true);
clo.Add<std::string>("method","lala");
}
then the program will fail to compile. On the other hand, changing the
first overload to:
template <typename ValueType>
void Add( std::string const& arg, bool required )
{ say( "bool" ); }
then the output will be "bool", as you expected.
On the third hand,
#include <iostream>
#include <ostream>
#include <string>
void say( char const s[] ) { std::cout << s << std::endl; }
class CommandLineOptions
{
public:
void Add( std::string const& arg, bool required )
{ say( "bool" ); }
template <typename ValueType>
void Add( std::string const& arg, ValueType const& aDefault )
{ say( "T" ); }
};
int main()
{
CommandLineOptions().Add( "method", "invert" );
CommandLineOptions().Add( "method", true );
CommandLineOptions().Add<std::string>( "method", "invert" );
CommandLineOptions().Add<bool>( "method", true );
}
produces "T bool T T", which is only undesired for the last call, and on
the fourth hand[1],
#include <iostream>
#include <ostream>
#include <string>
void say( char const s[] ) { std::cout << s << std::endl; }
class CommandLineOptions
{
public:
template <typename ValueType>
void Add( std::string const& arg, ValueType const& aDefault )
{ say( "T" ); }
};
template <>
void CommandLineOptions::Add<bool>(
std::string const& arg, bool const& required )
{ say( "bool" ); }
int main()
{
CommandLineOptions().Add( "method", "invert" );
CommandLineOptions().Add( "method", true );
CommandLineOptions().Add<std::string>( "method", "invert" );
CommandLineOptions().Add<bool>( "method", true );
}
produces "T bool T bool", sort of perfect.
Which also illustrates the difference between function overloading
(Tarjei's original code) and template specialization (this code[2]). :-)
Cheers,
- Alf
Notes:
[1] If 'clo.Add<std::string>( "method", true );' is to be supported then
we'd be on to the fifth hand, but I don't know whether there is one.
[2] This time I put the code through MSVC 7.1, g++ 3.4.4 and Comeau Online.
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