Re: Two Templates

"Jim Langston" <>
Sat, 23 Oct 2010 14:22:46 -0700
"KevinSimonson" <> wrote in message

Can someone tell me what the following code does? The first
"template<typename T, Endianess endianPolicy> struct TransformTo"
makes sense to me, but I don't understand what the "template<typename
T > struct TransformTo< T, Machine::endianess>" does. Also, in the
first one "endianPolicy" is mentioned but never used. So does that
mean including it has no effect? I'm kind of confused on that.

//! \brief Generic function object to give its char serialization a
//! specified byte ordering.
//! The byte ordering of the argument is swapped unless it matches the
//! ordering of the target machine.
//! We use partial specialization to achieve this.
template<typename T, Endianess endianPolicy> struct TransformTo
T operator()(T value) const { return swapEndianess< T >( value ); }
template<typename T > struct TransformTo< T, Machine::endianess >
T operator()(T value) const { return value; }

Kevin S

Somewhere machine::endianess is defined or typedefed or such to represent a
specific endian policy for that particular machine. The second one is
called a "Template specialiation". If the Endianess does not match
Machine::endianes then the first version of the template will be
called/built that returns the swapped values. However, if the endianPolicy
matches Machien::endianess, then the second template will be called/built
that simply returns the value. If the endianess already matches the
endianess of the machine do nothing.

I am curious, however, why this TransformTo even exists, why that logic
isn't being handled in the tempalte swapEndianess<T> which could do the
exact same thing if the endianess is the same, simply return the value. I
do not know of any other reasoning for specialiazing then leaving it alone,
and that is what swapEndianess<T> should do in the firstplace.

I'm curious what happens if you call swapEndianess< T > with a value with
the same endian policy. swapEndianess is probably interesting to look at

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