I am new to C++11. I am writing the following recursive lambda function, but it doesn't compile.
sum.cpp
#include <iostream>
#include <functional>
auto term = [](int a)->int {
return a*a;
};
auto next = [](int a)->int {
return ++a;
};
auto sum = [term,next,&sum](int a, int b)mutable ->int {
if(a>b)
return 0;
else
return term(a) + sum(next(a),b);
};
int main(){
std::cout<<sum(1,10)<<std::endl;
return 0;
}
compilation error:
vimal#linux-718q:~/Study/09C++/c++0x/lambda> g++ -std=c++0x sum.cpp
sum.cpp: In lambda function:
sum.cpp:18:36: error: ‘((<lambda(int, int)>*)this)-><lambda(int, int)>::sum’ cannot be used as a function
gcc version
gcc version 4.5.0 20091231 (experimental) (GCC)
But if I change the declaration of sum() as below, it works:
std::function<int(int,int)> sum = [term,next,&sum](int a, int b)->int {
if(a>b)
return 0;
else
return term(a) + sum(next(a),b);
};
Could someone please throw light on this?
Think about the difference between the auto version and the fully specified type version. The auto keyword infers its type from whatever it's initialized with, but what you're initializing it with needs to know what its type is (in this case, the lambda closure needs to know the types it's capturing). Something of a chicken-and-egg problem.
On the other hand, a fully specified function object's type doesn't need to "know" anything about what is being assigned to it, and so the lambda's closure can likewise be fully informed about the types its capturing.
Consider this slight modification of your code and it may make more sense:
std::function<int(int, int)> sum;
sum = [term, next, &sum](int a, int b) -> int {
if (a > b)
return 0;
else
return term(a) + sum(next(a), b);
};
Obviously, this wouldn't work with auto. Recursive lambda functions work perfectly well (at least they do in MSVC, where I have experience with them), it's just that they aren't really compatible with type inference.
The trick is to feed in the lambda implementation to itself as a parameter, not by capture.
const auto sum = [term, next](int a, int b) {
auto sum_impl = [term, next](int a, int b, auto& sum_ref) mutable {
if (a > b) {
return 0;
}
return term(a) + sum_ref(next(a), b, sum_ref);
};
return sum_impl(a, b, sum_impl);
};
All problems in computer science can be solved by another level of indirection. I first found this easy trick at http://pedromelendez.com/blog/2015/07/16/recursive-lambdas-in-c14/
It does require C++14 while the question is on C++11, but perhaps interesting to most.
Here's the full example at Godbolt.
Going via std::function is also possible but can result in slower code. But not always. Have a look at the answers to std::function vs template
This is not just a peculiarity about C++,
it's directly mapping to the mathematics of lambda calculus. From Wikipedia:
Lambda calculus cannot express this as directly as some other
notations:
all functions are anonymous in lambda calculus, so we can't refer to a
value which is yet to be defined, inside the lambda term defining that
same value. However, recursion can still be achieved by arranging for a
lambda expression to receive itself as its argument value
With C++14, it is now quite easy to make an efficient recursive lambda without having to incur the additional overhead of std::function, in just a few lines of code:
template <class F>
struct y_combinator {
F f; // the lambda will be stored here
// a forwarding operator():
template <class... Args>
decltype(auto) operator()(Args&&... args) const {
// we pass ourselves to f, then the arguments.
return f(*this, std::forward<Args>(args)...);
}
};
// helper function that deduces the type of the lambda:
template <class F>
y_combinator<std::decay_t<F>> make_y_combinator(F&& f) {
return {std::forward<F>(f)};
}
with which your original sum attempt becomes:
auto sum = make_y_combinator([term,next](auto sum, int a, int b) -> int {
if (a>b) {
return 0;
}
else {
return term(a) + sum(next(a),b);
}
});
In C++17, with CTAD, we can add a deduction guide:
template <class F> y_combinator(F) -> y_combinator<F>;
Which obviates the need for the helper function. We can just write y_combinator{[](auto self, ...){...}} directly.
In C++20, with CTAD for aggregates, the deduction guide won't be necessary.
In C++23, with deducing this, you don't need a Y-combinator at all:
auto sum = [term,next](this auto const& sum, int a, int b) -> int {
if (a>b) {
return 0;
}
else {
return term(a) + sum(next(a),b);
}
}
I have another solution, but work only with stateless lambdas:
void f()
{
static int (*self)(int) = [](int i)->int { return i>0 ? self(i-1)*i : 1; };
std::cout<<self(10);
}
Trick here is that lambdas can access static variables and you can convert stateless ones to function pointer.
You can use it with standard lambdas:
void g()
{
int sum;
auto rec = [&sum](int i) -> int
{
static int (*inner)(int&, int) = [](int& _sum, int i)->int
{
_sum += i;
return i>0 ? inner(_sum, i-1)*i : 1;
};
return inner(sum, i);
};
}
Its work in GCC 4.7
To make lambda recursive without using external classes and functions (like std::function or fixed-point combinator) one can use the following construction in C++14 (live example):
#include <utility>
#include <list>
#include <memory>
#include <iostream>
int main()
{
struct tree
{
int payload;
std::list< tree > children = {}; // std::list of incomplete type is allowed
};
std::size_t indent = 0;
// indication of result type here is essential
const auto print = [&] (const auto & self, const tree & node) -> void
{
std::cout << std::string(indent, ' ') << node.payload << '\n';
++indent;
for (const tree & t : node.children) {
self(self, t);
}
--indent;
};
print(print, {1, {{2, {{8}}}, {3, {{5, {{7}}}, {6}}}, {4}}});
}
prints:
1
2
8
3
5
7
6
4
Note, result type of lambda should be specified explicitly.
You can make a lambda function call itself recursively. The only thing you need to do is to is to reference it through a function wrapper so that the compiler knows it's return and argument type (you can't capture a variable -- the lambda itself -- that hasn't been defined yet).
function<int (int)> f;
f = [&f](int x) {
if (x == 0) return 0;
return x + f(x-1);
};
printf("%d\n", f(10));
Be very careful not to run out of the scope of the wrapper f.
I ran a benchmark comparing a recursive function vs a recursive lambda function using the std::function<> capture method. With full optimizations enabled on clang version 4.1, the lambda version ran significantly slower.
#include <iostream>
#include <functional>
#include <chrono>
uint64_t sum1(int n) {
return (n <= 1) ? 1 : n + sum1(n - 1);
}
std::function<uint64_t(int)> sum2 = [&] (int n) {
return (n <= 1) ? 1 : n + sum2(n - 1);
};
auto const ITERATIONS = 10000;
auto const DEPTH = 100000;
template <class Func, class Input>
void benchmark(Func&& func, Input&& input) {
auto t1 = std::chrono::high_resolution_clock::now();
for (auto i = 0; i != ITERATIONS; ++i) {
func(input);
}
auto t2 = std::chrono::high_resolution_clock::now();
auto duration = std::chrono::duration_cast<std::chrono::milliseconds>(t2-t1).count();
std::cout << "Duration: " << duration << std::endl;
}
int main() {
benchmark(sum1, DEPTH);
benchmark(sum2, DEPTH);
}
Produces results:
Duration: 0 // regular function
Duration: 4027 // lambda function
(Note: I also confirmed with a version that took the inputs from cin, so as to eliminate compile time evaluation)
Clang also produces a compiler warning:
main.cc:10:29: warning: variable 'sum2' is uninitialized when used within its own initialization [-Wuninitialized]
Which is expected, and safe, but should be noted.
It's great to have a solution in our toolbelts, but I think the language will need a better way to handle this case if performance is to be comparable to current methods.
Note:
As a commenter pointed out, it seems latest version of VC++ has found a way to optimize this to the point of equal performance. Maybe we don't need a better way to handle this, after all (except for syntactic sugar).
Also, as some other SO posts have outlined in recent weeks, the performance of std::function<> itself may be the cause of slowdown vs calling function directly, at least when the lambda capture is too large to fit into some library-optimized space std::function uses for small-functors (I guess kinda like the various short string optimizations?).
Here is a refined version of the Y-combinator solution based on one proposed by #Barry.
template <class F>
struct recursive {
F f;
template <class... Ts>
decltype(auto) operator()(Ts&&... ts) const { return f(std::ref(*this), std::forward<Ts>(ts)...); }
template <class... Ts>
decltype(auto) operator()(Ts&&... ts) { return f(std::ref(*this), std::forward<Ts>(ts)...); }
};
template <class F> recursive(F) -> recursive<F>;
auto const rec = [](auto f){ return recursive{std::move(f)}; };
To use this, one could do the following
auto fib = rec([&](auto&& fib, int i) {
// implementation detail omitted.
});
It is similar to the let rec keyword in OCaml, although not the same.
In C++23 deducing this (P0847) will be added:
auto f = [](this auto& self, int i) -> int
{
return i > 0 ? self(i - 1) + i : 0;
}
For now its only available in EDG eccp and (partially) available in MSVC:
https://godbolt.org/z/f3E3xT3fY
This is a slightly simpler implementation of the fixpoint operator which makes it a little more obvious exactly what's going on.
#include <iostream>
#include <functional>
using namespace std;
template<typename T, typename... Args>
struct fixpoint
{
typedef function<T(Args...)> effective_type;
typedef function<T(const effective_type&, Args...)> function_type;
function_type f_nonr;
T operator()(Args... args) const
{
return f_nonr(*this, args...);
}
fixpoint(const function_type& p_f)
: f_nonr(p_f)
{
}
};
int main()
{
auto fib_nonr = [](const function<int(int)>& f, int n) -> int
{
return n < 2 ? n : f(n-1) + f(n-2);
};
auto fib = fixpoint<int,int>(fib_nonr);
for (int i = 0; i < 6; ++i)
{
cout << fib(i) << '\n';
}
}
C++ 14:
Here is a recursive anonymous stateless/no capture generic set of lambdas
that outputs all numbers from 1, 20
([](auto f, auto n, auto m) {
f(f, n, m);
})(
[](auto f, auto n, auto m) -> void
{
cout << typeid(n).name() << el;
cout << n << el;
if (n<m)
f(f, ++n, m);
},
1, 20);
If I understand correctly this is using the Y-combinator solution
And here is the sum(n, m) version
auto sum = [](auto n, auto m) {
return ([](auto f, auto n, auto m) {
int res = f(f, n, m);
return res;
})(
[](auto f, auto n, auto m) -> int
{
if (n > m)
return 0;
else {
int sum = n + f(f, n + 1, m);
return sum;
}
},
n, m); };
auto result = sum(1, 10); //result == 55
Here's the proof that a recursive lambda with a small body almost has the same performance like a usual recursive fuction which can call itself directly.
#include <iostream>
#include <chrono>
#include <type_traits>
#include <functional>
#include <atomic>
#include <cmath>
using namespace std;
using namespace chrono;
unsigned recursiveFn( unsigned x )
{
if( x ) [[likely]]
return recursiveFn( x - 1 ) + recursiveFn( x - 1 );
else
return 0;
};
atomic_uint result;
int main()
{
auto perf = []( function<void ()> fn ) -> double
{
using dur_t = high_resolution_clock::duration;
using urep_t = make_unsigned_t<dur_t::rep>;
high_resolution_clock::duration durMin( (urep_t)-1 >> 1 );
for( unsigned r = 10; r--; )
{
auto start = high_resolution_clock::now();
fn();
dur_t dur = high_resolution_clock::now() - start;
if( dur < durMin )
durMin = dur;
}
return durMin.count() / 1.0e9;
};
auto recursiveLamdba = []( auto &self, unsigned x ) -> unsigned
{
if( x ) [[likely]]
return self( self, x - 1 ) + self( self, x - 1 );
else
return 0;
};
constexpr unsigned DEPTH = 28;
double
tLambda = perf( [&]() { ::result = recursiveLamdba( recursiveLamdba, DEPTH ); } ),
tFn = perf( [&]() { ::result = recursiveFn( DEPTH ); } );
cout << trunc( 1000.0 * (tLambda / tFn - 1.0) + 0.5 ) / 10.0 << "%" << endl;
}
For my AMD Zen1 CPU with current MSVC the recursiveFn is about 10% faster. For my Phenom II x4 945 with g++ 11.1.x both functions have the same performance.
Keep in mind that this is almost the worst case since the body of the funtion is very small. If it is larger the part of the recursive function call itself is smaller.
You're trying to capture a variable (sum) you're in the middle of defining. That can't be good.
I don't think truely self-recursive C++0x lambdas are possible. You should be able to capture other lambdas, though.
Here is the final answer for the OP. Anyway, Visual Studio 2010 does not support capturing global variables. And you do not need to capture them because global variable is accessable globally by define. The following answer uses local variable instead.
#include <functional>
#include <iostream>
template<typename T>
struct t2t
{
typedef T t;
};
template<typename R, typename V1, typename V2>
struct fixpoint
{
typedef std::function<R (V1, V2)> func_t;
typedef std::function<func_t (func_t)> tfunc_t;
typedef std::function<func_t (tfunc_t)> yfunc_t;
class loopfunc_t {
public:
func_t operator()(loopfunc_t v)const {
return func(v);
}
template<typename L>
loopfunc_t(const L &l):func(l){}
typedef V1 Parameter1_t;
typedef V2 Parameter2_t;
private:
std::function<func_t (loopfunc_t)> func;
};
static yfunc_t fix;
};
template<typename R, typename V1, typename V2>
typename fixpoint<R, V1, V2>::yfunc_t fixpoint<R, V1, V2>::fix = [](tfunc_t f) -> func_t {
return [f](fixpoint<R, V1, V2>::loopfunc_t x){ return f(x(x)); }
([f](fixpoint<R, V1, V2>::loopfunc_t x) -> fixpoint<R, V1, V2>::func_t{
auto &ff = f;
return [ff, x](t2t<decltype(x)>::t::Parameter1_t v1,
t2t<decltype(x)>::t::Parameter1_t v2){
return ff(x(x))(v1, v2);
};
});
};
int _tmain(int argc, _TCHAR* argv[])
{
auto term = [](int a)->int {
return a*a;
};
auto next = [](int a)->int {
return ++a;
};
auto sum = fixpoint<int, int, int>::fix(
[term,next](std::function<int (int, int)> sum1) -> std::function<int (int, int)>{
auto &term1 = term;
auto &next1 = next;
return [term1, next1, sum1](int a, int b)mutable ->int {
if(a>b)
return 0;
else
return term1(a) + sum1(next1(a),b);
};
});
std::cout<<sum(1,10)<<std::endl; //385
return 0;
}
This answer is inferior to Yankes' one, but still, here it goes:
using dp_type = void (*)();
using fp_type = void (*)(dp_type, unsigned, unsigned);
fp_type fp = [](dp_type dp, unsigned const a, unsigned const b) {
::std::cout << a << ::std::endl;
return reinterpret_cast<fp_type>(dp)(dp, b, a + b);
};
fp(reinterpret_cast<dp_type>(fp), 0, 1);
You need a fixed point combinator. See this.
or look at the following code:
//As decltype(variable)::member_name is invalid currently,
//the following template is a workaround.
//Usage: t2t<decltype(variable)>::t::member_name
template<typename T>
struct t2t
{
typedef T t;
};
template<typename R, typename V>
struct fixpoint
{
typedef std::function<R (V)> func_t;
typedef std::function<func_t (func_t)> tfunc_t;
typedef std::function<func_t (tfunc_t)> yfunc_t;
class loopfunc_t {
public:
func_t operator()(loopfunc_t v)const {
return func(v);
}
template<typename L>
loopfunc_t(const L &l):func(l){}
typedef V Parameter_t;
private:
std::function<func_t (loopfunc_t)> func;
};
static yfunc_t fix;
};
template<typename R, typename V>
typename fixpoint<R, V>::yfunc_t fixpoint<R, V>::fix =
[](fixpoint<R, V>::tfunc_t f) -> fixpoint<R, V>::func_t {
fixpoint<R, V>::loopfunc_t l = [f](fixpoint<R, V>::loopfunc_t x) ->
fixpoint<R, V>::func_t{
//f cannot be captured since it is not a local variable
//of this scope. We need a new reference to it.
auto &ff = f;
//We need struct t2t because template parameter
//V is not accessable in this level.
return [ff, x](t2t<decltype(x)>::t::Parameter_t v){
return ff(x(x))(v);
};
};
return l(l);
};
int _tmain(int argc, _TCHAR* argv[])
{
int v = 0;
std::function<int (int)> fac =
fixpoint<int, int>::fix([](std::function<int (int)> f)
-> std::function<int (int)>{
return [f](int i) -> int{
if(i==0) return 1;
else return i * f(i-1);
};
});
int i = fac(10);
std::cout << i; //3628800
return 0;
}
Related
I'm new to C++ 14, I want to compose a variable length of lambda functions together to a single lambda, how should I do that? Below is my current work
#include <iostream>
template<typename Ftype>
Ftype compose(const Ftype & fn) {
return fn;
}
template<typename Ftype, typename... Other>
auto compose(Ftype fn, Other... other) {
return [=](auto x){return other(fn(x))...;};
} ➤ expression contains unexpanded parameter pack 'other'
int main(void) {
auto add_func = [](const int x) { return x * 7; };
auto sub_func = [](const int x) { return x + 1; };
int res = compose(add_func, sub_func)(1);
std::cout << "Result: " << res << "\n";
}
But I failed the compile, I guess I may use the lambda or variadic parameter somehow wrong.
Can someone help me?
Your "recursive" case contains no call to compose, that should be a hint of where you've mixed things up ;)
return [=](auto x){ return compose(other...)(fn(x)); };
Suppose I have a naive implementation of an applicative, which is a name I picked for sanity and not that I know anything about Applicative type class from other languages. Here goes the implementation:
#include <iostream>
#include <string>
template <typename T>
struct applicative {
template <typename Fn>
auto then(Fn f) const {
return applicative<decltype(f(data_))>{f(data_)};
}
template <typename Fn>
auto and_last(Fn f) const {
return f(data_);
}
T data_;
};
int main() {
applicative<std::string>{"hello world"}
.then([](std::string const& s) {return s.size() * 4; })
.then([](int k) {return k - 2; })
.and_last([](int k) { std::cout << k << "\n"; });
}
Now, this could be improved in many ways. Providing something like make_applicative for in-place construction, trying to eliminate redundant copies and moves if there are any etc. But from the time I saw our ability to compose functions in C++, I feel like something better is possible. Any compose implementation would work, so let's pick the one in this codereview.stackexchange question. With it, we can say things like
auto f1 = [](std::pair<double,double> p) {return p.first + p.second; };
auto f2 = [](double x) {return std::make_pair(x, x + 1.0); };
auto f3 = [](double x, double y) {return x*y; };
auto g = compose(f1, f2, f3);
std::cout << g(2.0, 3.0) << std::endl; //prints '13', evaluated as (2*3) + ((2*3)+1)
Pretty nice. Going back to my idea, I think this should make possible a rework of my applicative implementation in following way:
auto sf = applicative<std::string>{}
.then([](std::string const& s) {return s.size() * 4; })
.then([](int k) {return k - 2; });
std::cout << sf.eval_with("hello world"); << "\n";
As you can see, this is sort-of lazy evaluated and we only supply value when we need it, with eval_with. I've been thinking about how to implement this new version for one hour now and I have no idea where to store operations and composed functions, what to make of applicative template parameter like we have here with std::string and many more problems. How would one implement something like this? Is it trivial as I initially hoped to be or does it require a lot of code? I really want this because I feel like this would buy me a lot by preventing a lot of argument passing on long chain of functions.
Edit: I am working on it a little more and turns out compose implementation I linked is not what I actually had in mind since we are still performing all the function calls in the chain and still passing arguments around. But you can answer assuming any compose function would work and choice of a better compose implementation would be a performance optimization. What I had in mind was more like following, taken from my example
applicative<std::string>{"hello world"}
.then([](std::string const& s) {return s.size() * 4; })
.then([](int k) {return k - 2; })
.and_last([](int k) { std::cout << k << "\n"; });
Those then calls would result in a single function call equivalent to (s.size() * 4) - 2 which could be evaluated with eval_with.
Is that what you want?
#include <iostream>
#include <string>
struct id
{
template <typename T>
auto operator()(T t) const
{
return t;
}
};
template <typename T, typename Func = id>
struct applicative {
applicative(Func f = Func())
: _f(f)
{
}
template <typename Fn>
auto then(Fn f) const {
auto composition = [=](T val) { return f(_f(val)); };
return applicative<T, decltype(composition)>(composition);
}
auto eval_with(T t)
{
return _f(t);
}
Func _f;
};
int main() {
auto sf = applicative<std::string>{}
.then([](std::string const& s) {return s.size() * 4; })
.then([](int k) {return k - 2; });
std::cout << sf.eval_with("hello world") << "\n";
}
Disclaimer: I didn't bother about perfect forwarding, so everything is passed by value.
Is it possible to evaluate std::optional::value_or(expr) argument in a lazy way, so the expr were calculated only in the case of having no value?
If not, what would be a proper replacement?
#include <optional>
template <typename F>
struct Lazy
{
F f;
operator decltype(f())() const
{
return f();
}
};
template <typename F>
Lazy(F f) -> Lazy<F>;
int main()
{
std::optional<int> o;
int i = o.value_or(Lazy{[]{return 0;}});
}
DEMO
You may write your helper function:
template<typename T, typename F>
T lazy_value_or(const std::optional<T> &opt, F fn) {
if(opt) return opt.value();
return fn();
}
which can then be used as:
T t = lazy_value_or(opt, [] { return expensive_computation();});
If that's significantly less typing than doing it explicitly that's up to you to judge; still, you can make it shorter with a macro:
#define LAZY_VALUE_OR(opt, expr) \
lazy_value_or((opt), [&] { return (expr);})
to be used as
T t = LAZY_VALUE_OR(opt, expensive_calculation());
This is closest to what I think you want, but may be frowned upon as it hides a bit too much stuff.
Make an optional of function type.
Then a lambda can be passed in, which when called will calculate the correct value at the requested moment.
std::optional<std::function<int()>> opt;
int a = 42;
opt = [=] { return a; }
int b = 4;
int c = opt.value_or([=] { return b * 10 + 2;}) ();
I'm trying to create a bitset according to the type send to the function. But let's reduce the test case a little.
Warning : I'm using auto gcc extension for this example, I don't need to use template parameter.
namespace hana = boost::hana;
constexpr decltype(auto) rec(auto i, auto max, auto f, auto returnValue) {
return returnValue |= f(i);
if constexpr (i < max) //"infinite" loop if no constexpr
return rec(i + hana::size_c<1>, max, f, returnValue);
else
return returnValue;
}
constexpr decltype(auto) foo(auto ct, auto ... type) {
constexpr auto tuple = hana::make_tuple(type...);
constexpr unsigned long returnValue = 0L;
constexpr auto f = [tuple, ct] (auto i) {
if (hana::contains(tuple, ct[i]))
return 0 << decltype(i)::value;
else
return 0;
};
return rec(hana::size_c<0>, hana::size_c<3>, f, returnValue);
}
struct T1 {};
struct T2 {};
struct T3 {};
int main () {
constexpr auto t1 = hana::type_c<T1>;
constexpr auto t2 = hana::type_c<T2>;
constexpr auto t3 = hana::type_c<T3>;
constexpr auto ct = hana::make_tuple(t1, t2, t3);
constexpr auto test = foo(ct, t1, t2);
}
Seems like my tuple is not considered Searchable, but if I try the same hana::contains outside the lambda I got no problem.
The whole error is huge so check it there : live demo
By the way, I tried to do this with a for loop but failed. Do you know a good way of doing this kind of things in C++17/20 ?
The error is caused by an out of bounds access caused by the use of manual recursion. Part of the purpose of functional programming is to provide constructs to eliminate the possibility of these kinds of mistakes.
Here are a few examples, but it is recommended to take a look at the manual for the concept hana::Foldable as it is really foundational in using Boost.Hana.
hana::fold_left hides the recursion for you and can reduce the amount of recursive calls via fast-tracking:
constexpr decltype(auto) foo = [](auto ct, auto ... type) {
constexpr auto tuple = hana::make_tuple(type...);
return hana::fold_left(hana::make_range(hana::size_c<0>, hana::size_c<3>), 0L,
[tuple, ct](auto returnValue, auto i)
{
// returnValue param is not constexpr
if (hana::contains(tuple, ct[i])) {
return returnValue | (1 << decltype(i)::value);
}
else
{
return returnValue;
}
}
);
};
hana::fold_left example
hana::unpack eliminates recursion altogether using variadic pack expansion:
constexpr decltype(auto) foo = [](auto ct, auto ... type) {
constexpr auto tuple = hana::make_tuple(type...);
auto f = [tuple, ct](auto i)
{
return hana::contains(tuple, ct[i]) ? (1 << decltype(i)::value) : 0;
};
return hana::unpack(hana::make_range(hana::size_c<0>, hana::size_c<3>),
[f](auto ...i) { return (f(i) | ...); }
);
};
hana::unpack example
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.
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.