Re: cloning a basic type efficiently
terry wrote:
Hi, if one wants to create a new type that behaves in all practical ways
like the original basic type except regarding type conversions, is there a
better way to do this than:
class my_double
{
public:
my_double(void) : _val(0){};
You forgot to declare _val, which I suppose is a private double, member
of my_double.
There is no need to write "void". This is the default, unlike C.
There is no need for semicolons at the end of function definitions.
my_double( const my_double & rhs) : _val(rhs._val){};
This is already defined by default.
explicit my_double( const double & rhs) : _val(rhs){};
People usually say that there is no need to take a native type by const
reference.
But who knows, doing so might prevent optimizations.
~my_double(void){};
my_double operator+ ( const my_double& rhs )
{ return my_double(_val * rhs._val;)
operator+ is usually a free function, and usually implemented on top of
operator+= (which is a member).
You may consider returning a const my_double also, unless you want to allow
my_double(6.) + my_double(3.) = my_double(5.);
Also, usually operator+ performs addition, not multiplication. I suppose
this is a typo.
I find, in order, the following issues troublesome:
1) performance - can I make expect compilers to not introduce extra
processing and lose performance.
A conforming compiler does not have to make it as efficient.
It could introduce pointers for every reference.
But if it's smart enough and inline the whole thing, that should be
quite equivalent to using doubles directly.
2) clarity and error - having to do something for each memember function
separately is not a good idea
You can't expect a C++ compiler to guess what your binary operator* is
supposed to do, even if it has a binary operator+ defined.
I really don't understand what you mean here.
3) cloning classes in C++ is troublesome
Looks like you haven't really programmed with C++ yet. This is the very
basics.
It's not cloning, actually, but copying. Cloning is copying polymorphically.
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