在C ++ 11中编写通用memoization函数 [英] Writing Universal memoization function in C++11

问题描述

寻找一种方法来实现一个通用的通用memoization函数，它将采用一个函数并返回相同的记忆版本？

Looking for a way to implement a universal generic memoization function which will take a function and return the memoized version of the same?

寻找类似@memo来自Norving的网站）在python中的装饰器。

Looking for something like @memo (from Norving's site)decorator in python.

def memo(f):
    table = {}
    def fmemo(*args):
        if args not in table:
            table[args] = f(*args)
        return table[args]
    fmemo.memo = table
    return fmemo

在C ++中表达通用和可重用的装饰器，可能使用C ++ 11的新特性？

Going more general, is there a way to express generic and reusable decorators in C++, possibly using the new features of C++11?

推荐答案

：

template <typename R, typename... Args>
std::function<R (Args...)> memo(R (*fn)(Args...)) {
    std::map<std::tuple<Args...>, R> table;
    return [fn, table](Args... args) mutable -> R {
        auto argt = std::make_tuple(args...);
        auto memoized = table.find(argt);
        if(memoized == table.end()) {
            auto result = fn(args...);
            table[argt] = result;
            return result;
        } else {
            return memoized->second;
        }
    };
}

在C ++ 14中，可以使用广义返回类型通过返回 std :: function 强加的额外间接。

In C++14, one can use generalized return type deduction to avoid the extra indirection imposed by returning std::function.

他们在 std :: function 首先留给读者做练习。

Making this fully general, permitting passing arbitrary function objects without wrapping them in std::function first is left as an exercise for the reader.

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