Default function that just returns the passed valu

2019-02-06 06:22发布

As a lazy developer, I like to use this trick to specify a default function:

template <class Type, unsigned int Size, class Function = std::less<Type> >
void arrange(std::array<Type, Size> &x, Function&& f = Function())
{
    std::sort(std::begin(x), std::end(x), f);
}

But I have a problem in a very particular case, which is the following:

template <class Type, unsigned int Size, class Function = /*SOMETHING 1*/>
void index(std::array<Type, Size> &x, Function&& f = /*SOMETHING 2*/)
{
    for (unsigned int i = 0; i < Size; ++i) {
        x[i] = f(i);
    }
}

In this case, I would like the default function to be the equivalent of: [](const unsigned int i){return i;} (a function that just returns the passed value).

In order to do that, what do I have to write instead of /*SOMETHING 1*/ and /*SOMETHING 2*/?

5条回答
疯言疯语
2楼-- · 2019-02-06 06:35

There is no standard functor that does this, but it is easy enough to write (though the exact form is up for some dispute):

struct identity {
    template<typename U>
    constexpr auto operator()(U&& v) const noexcept
        -> decltype(std::forward<U>(v))
    {
        return std::forward<U>(v);
    }
};

This can be used as follows:

template <class Type, std::size_t Size, class Function = identity>
void index(std::array<Type, Size> &x, Function&& f = Function())
{
    for (unsigned int i = 0; i < Size; ++i) {
        x[i] = f(i);
    }
}
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戒情不戒烟
3楼-- · 2019-02-06 06:44

You can just build your own identity functor:

template <typename T>
class returnIdentifyFunctor
{
  public:
     auto operator ()(  T &&i ) -> decltype( std::forward<T>(i) )
    {
      return std::move(i);
    }
};

template <class Type, unsigned int Size, class Function = returnIdentifyFunctor<Type>>
void index(std::array<Type, Size> &x, Function&& f = Function() )
 {
    for (unsigned int i = 0; i < Size; ++i) {
            x[i] = f(i);
  }
}
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欢心
4楼-- · 2019-02-06 06:46

boost::phoenix offers a complete functional toolbox, here 'arg1' is the ident to identity ;-)

#include <boost/phoenix/core.hpp>

template <class X, class Function = decltype(boost::phoenix::arg_names::arg1)>
void index(X &x, Function f = Function()) {
    for (std::size_t i = 0; i < x.size(); ++i) {
            x[i] = f(i);
  }
}
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兄弟一词,经得起流年.
5楼-- · 2019-02-06 06:46

This is called the identity function. Unfortunately, it is not part of the C++ standard, but you can easily build one yourself.


If you happen to use g++, you can activate its extensions with -std=gnu++11 and then

#include <array>
#include <ext/functional>

template <class Type, std::size_t Size, class Function = __gnu_cxx::identity<Type> >
void index(std::array<Type, Size> &x, Function&& f = Function())
{
    for (unsigned int i = 0; i < Size; ++i) {
        x[i] = f(i);
    }
}

Maybe it will be available in C++20, see std::identity. Until then you may look at boost's version at boost::compute::identity.

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你好瞎i
6楼-- · 2019-02-06 06:52

A way to deal with this is to have two different functions. I find quite sane not to use default parameters.

template <class Type, unsigned int Size, class Function>
void index(std::array<Type, Size> &x, Function&& f){
    for(unsigned int i = 0; i < Size; ++i) x[i] = f(i);
}

template<class Type, unsigned int Size>
void index(std::array<Type, Size> &x){
    return index(x, [](unsigned int i){return i;});                      // C++11 in this case
                //, [](auto&& e){return std::forward<decltype(e)>(e)};); // C++14 in a more general case
                //, std::identity); // C++20 in general
}
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