Good day,
A friend of mine is asking about transforming an integer square root function into a meta-function. Here is the original function:
unsigned isqrt(unsigned value)
{
unsigned sq = 1, dlt = 3;
while(sq<=value)
{
sq += dlt;
dlt += 2;
}
return (dlt>>1) - 1;
}
I wrote a meta version using constexpr
, but he said he can't use the new feature for some reason:
constexpr std::size_t isqrt_impl
(std::size_t sq, std::size_t dlt, std::size_t value){
return sq <= value ?
isqrt_impl(sq+dlt, dlt+2, value) : (dlt >> 1) - 1;
}
constexpr std::size_t isqrt(std::size_t value){
return isqrt_impl(1, 3, value);
}
So I thought it shouldn't be that hard to transform that into template struct that calls it self recursively:
template <std::size_t value, std::size_t sq, std::size_t dlt>
struct isqrt_impl{
static const std::size_t square_root =
sq <= value ?
isqrt_impl<value, sq+dlt, dlt+2>::square_root :
(dlt >> 1) - 1;
};
template <std::size_t value>
struct isqrt{
static const std::size_t square_root =
isqrt_impl<value, 1, 3>::square_root;
};
Unfortunately, this is causing infinite recursion(on GCC 4.6.1) and I am unable to figure out what is wrong with the code. Here is the error:
C:\test>g++ -Wall test.cpp
test.cpp:6:119: error: template instantiation depth exceeds maximum of 1024 (use
-ftemplate-depth= to increase the maximum) instantiating 'struct isqrt_impl<25u
, 1048576u, 2049u>'
test.cpp:6:119: recursively instantiated from 'const size_t isqrt_impl<25u, 4u
, 5u>::square_root'
test.cpp:6:119: instantiated from 'const size_t isqrt_impl<25u, 1u, 3u>::squar
e_root'
test.cpp:11:69: instantiated from 'const size_t isqrt<25u>::square_root'
test.cpp:15:29: instantiated from here
test.cpp:6:119: error: incomplete type 'isqrt_impl<25u, 1048576u, 2049u>' used i
n nested name specifier
Thanks all,
Unfortunately, this is causing infinite recursion (on GCC 4.6.1) and I am unable to figure out what is wrong with the code.
I don't see a base case specialization for isqrt_impl
. You need to have a template specialization for the base case to break this recursion. Here is a simple attempt at that:
template <std::size_t value, std::size_t sq, std::size_t dlt, bool less_or_equal = sq <= value >
struct isqrt_impl;
template <std::size_t value, std::size_t sq, std::size_t dlt>
struct isqrt_impl< value, sq, dlt, true >{
static const std::size_t square_root =
isqrt_impl<value, sq+dlt, dlt+2>::square_root;
};
template <std::size_t value, std::size_t sq, std::size_t dlt>
struct isqrt_impl< value, sq, dlt, false >{
static const std::size_t square_root =
(dlt >> 1) - 1;
};
Template evaluation isn't lazy by default.
static const std::size_t square_root =
sq <= value ?
isqrt_impl<value, sq+dlt, dlt+2>::square_root :
(dlt >> 1) - 1;
will always instantiate the template, no matter what the condition. You need boost::mpl::eval_if
or something equivalent to get that solution to work.
Alternatively you can have a base case partial template specialization that stops the recursion if the condition is met, like in K-ballos answer.
I'd actually prefer code that uses some form of lazy evaluation over partial specialization because I feel it is easier to comprehend and keeps the noise that comes with templates lower.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With