Re: templates in parameters
On Aug 1, 2:25 pm, ssylee <staniga...@gmail.com> wrote:
On Aug 1, 11:14 am, mlimber <mlim...@gmail.com> wrote:
On Aug 1, 1:53 pm, ssylee <staniga...@gmail.com> wrote:
I'm trying to template a function as follows in the pastebin
link(http://pastebin.com/m5e642c72).
Would I call the function like as follows:
MyType p = ConvFunc <RECT> (rectangle);
when rectangle is of type RECT?
The code is:
struct MyType { /*...*/ };
template <class T>
MyType ConvFunc(T input)
{
// ...
}
You don't need to specify the type of a parameter (unless you perhaps
wanted a derived class to be converted to its base class or something)
since the compiler can deduce it automatically. If the return value
were customizable, then you would need to specify that since it is not
a function argument and therefore not deducible:
template<class Ret, class Arg>
Ret Conv( const Arg& arg )
{
// ...
}
void Foo( const int n )
{
MyType m1 = Conv( n ); // Error! Can't figure out return ty=
pe
MyType m2 = Conv<MyType>( n ); // Ok
MyType m3 = Conv<MyType,int>( n ); // Ok but redundant
MyType m4 = Conv<MyType,double>( n ); // Ok, forces conversion
// ...
}
Thanks. I think you misunderstood what I'm trying to do. I'm trying to
convert the parameter (which could be in many different types) into
MyType. Would I need to use the power of templates to achieve the goal?
I don't believe I misunderstood you. I answered your question in the
sentence after the first code block and then attempted to demonstrate
a more general case where a function Conv() could convert between any
objects. Note also that you could use an overloaded cast operator to
do a conversion if the result type is a class:
#include <sstream>
class A
{
int i_;
public:
// ...
// Convert an A to a string automatically
operator std::string() const
{
std::ostringstream oss;
oss << i_;
}
};
void Foo( const A& a )
{
// These all do the same thing
std::string s1 = a; // implicitly calls A::operator std::string()
std::string s2 = static_cast<std::string>( a ); // Explicit
std::string s3 = Conv<std::string>( a ); // Use the func above
// ...
}
Cheers! --M