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Memoization functor wrapper in c++

Here is a generic memoization wrapper I wrote for functions. It makes use of tuplehash.

template<typename R, typename... Args>
class memofunc{
    typedef R (*func)(Args...);
    func fun_;
    unordered_map<tuple<Args...>, R, tuplehash::hash<tuple<Args...> > > map_;
public:
    memofunc(func fu):fun_(fu){}
    R operator()(Args&&... args){
    auto key = make_tuple(std::forward<Args>(args)...);
    auto q = map_.find(key);
    if(q == map_.end()){
        R res = fun_(std::forward<Args>(args)...);
        map_.insert({key,res});
        return res;
    }else{
        return q->second;
    }
    }
};

example of usage for Fibonacci numbers.

long long fibo(long long x){
    static memofunc<long long, long long> memf(fibo);
    // try to replace fibo with this new fibo but doesn't work, why?
    // function<long long(long long)> fibo = memf; 

    if(x <= 2) return 1;
    // this works but involves changing the original code.
    // how to write code such that I dont need to manually add this code in?
    return memf(x-1) + memf(x-2); 
    // old code
    // return fibo(x-1) + fibo(x-2);
}

Question is, ideally I could just add a few line to the beginning of the recursive function and done with memoization. But simple replacement doesn't work, and this is where I stuck.

like image 229
user40129 Avatar asked Jul 08 '14 17:07

user40129


1 Answers

Your problem seems to be that you make a local copy of your memoizer at each function call, then destroy it.

Here is a simple one-argument version of your memoizer that seems to work:

#include <iostream>
#include <functional>
#include <unordered_map>

template<typename Sig, typename F=Sig* >
struct memoize_t;
template<typename R, typename Arg, typename F>
struct memoize_t<R(Arg), F> {
  F f;
  mutable std::unordered_map< Arg, R > results;
  template<typename... Args>
  R operator()( Args&&... args ) const {
    Arg a{ std::forward<Args>(args)... }; // in tuple version, std::tuple<...> a
    auto it = results.find(a);
    if (it != results.end())
      return it->second;
    R retval = f(a); // in tuple version, use a tuple-to-arg invoker
    results.emplace( std::forward<Arg>(a), retval ); // not sure what to do here in tuple version
    return retval;
  }
};

template<typename F>
memoize_t<F> memoize( F* func ) {
  return {func};
}

int foo(int x) {
  static auto mem = memoize(foo);
  auto&& foo = mem;
  std::cout << "processing...\n";
  if (x <= 0) return foo(x+2)-foo(x+1); // bwahaha
  if (x <= 2) return 1;
  return foo(x-1) + foo(x-2);;
}
int main() {
  std::cout << foo(10) << "\n";
}

live example

Note that foo(10) only does 10 invocations of foo.

This also admits:

#define CAT2(A,B,C) A##B##C
#define CAT(A,B,C) CAT2(A,B,C)
#define MEMOIZE(F) \
  static auto CAT( memoize_static_, __LINE__, F ) = memoize(F); \
  auto&& F = CAT( memoize_static_, __LINE__, F )

int foo(int x) {
  MEMOIZE(foo);
  std::cout << "processing...\n";
  if (x <= 0) return 0;
  if (x <= 2) return 1;
  return foo(x-1) + foo(x-2);;
}

for people who like macros for this kind of thing.

A 3 step version might be better.

First, a prelude with a forward declaration of the function and memoizer wrapper.

Second, within the function, an alias for the function name, so recursive calls use the memorization function.

Third, after the declaration of the function, an alias for the function name, so external calls also use the memoized version.

The code above only memoizes recursive calls, never the initial call.

like image 164
Yakk - Adam Nevraumont Avatar answered Oct 29 '22 18:10

Yakk - Adam Nevraumont