Re: how to implement sum of bits using template metaprogramming
Alf P. Steinbach wrote:
* andrey.vul@gmail.com:
I'm tryin to convert this:
long sumOfBits(int n) {
long x = 0;
int i;
for (i = n - 1; i >= 0; i--)
x += (1 << i);
return x;
}
into something like this:
template <int n> struct sumOfBits {
static const unsigned char u8value = (unsigned char)((1 << n) +
sumOfBits<n - 1>::u8value);
static const unsigned __int16 u16value = (unsigned __int16)((1 << n)
+ sumOfBits<n - 1>::u16value);
static const unsigned __int32 u32value = (unsigned __int32)((1 << n)
+ sumOfBits<n - 1>::u32value);
static const unsigned __int64 u64value = (unsigned __int64)((1 << n)
+ sumOfBits<n - 1>::u64value);
};
template <> struct sumOfBits <0> {
static const unsigned char u8value = (unsigned char)1;
static const unsigned __int16 u16value = (unsigned __int16)1;
static const unsigned __int32 u32value = (unsigned __int32)1;
static const unsigned __int64 u64value = (unsigned __int64)1;
};
and so far, all expansions, that have multiples of 8 (8, 16, 32, 64)
return 0 due to overflow.
Any way to correctly rewrite the template?
Off the cuff, to reproduce the behavior of the original non-tempalte
function:
template< int n >
struct SumOfBits
{
enum{ value = (1L << (n-1)) | ((1L << (n-1)) - 1) };
};
I believe you meant
template< unsigned n > struct SumOfBits
{
enum{ value = (1L << (n - 1)) + SumOfBits<n-1>::value };
};
template<> struct SumOfBits<1>
{
enum { value = 1 };
};
template<> struct SumOfBits<0>
{
enum { value = 0 };
};
You need recursion in templates to emulate the loop.
V
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