Pass a string Recursively without Recreation

老子叫甜甜 提交于 2019-11-26 23:45:41

问题


I answered a question here: https://stackoverflow.com/a/28862668/2642059 Where I needed to use recurrence to step through a string. I wanted to use a const string& as my parameter on each function, but unless I wanted to reconstruct the string each recursion I found that I needed to pass a start and finish position as well as the string itself. So it became pointless to pass the string at all.

In the end I choose to just pass a start and finish pointer to the char[].


As an example, say that I'm given a string which contains nested parenthesis (but no side by side parenthetical insertions.) So like this:

(abc(def(ghi((j)klm)nop)qrs)tuv)wxyz

But not like this:

(abc(def)(ghi)(j)(klm)(nop)(qrs)tuv)wxyz

I want to write a recursive program to extract the string in the deepest nested parentheses. Something like:

string foo(const string& bar){
    auto start = bar.find('(') + 1;

    return start == string::npos + 1 ? bar : foo(bar.substr(start, bar.find_last_of(')') - start));
}

However I'm unhappy reconstructing a string for each recurrence of foo. The alternative is to pass start and finish pointers as in the linked example (or to pass string::const_iterators.)

Is there a wrapper or something which would allow me to use string functionality, but not reconstruct a string?


回答1:


string_view from the library fundamentals TS might be one idea, support is available in GCC.

The interface is virtually identical to string

#include <experimental/string_view>
using std::experimental::string_view;

string_view foo(const string_view& bar){
    auto start = bar.find('(') + 1;

    return start == string_view::npos + 1 ? bar : foo(bar.substr(start, bar.find_last_of(')') - start));
}

The last line could also be

return start ? foo(bar.substr(start, bar.find_last_of(')') - start)) : bar;

Although they're both pretty cryptic.




回答2:


Write your own array_view<T>. It is a few dozen lines of code.

Use std::find to replace both algorithms. In one case, use reverse iterators. (or write a range-based find and range-based backwards)

Use {T*,T*} ctor to recurse.

array_view<const char> foo(array_view<const char> bar)

Here is a primitive array_view<T>:

template<class T>
struct array_view {
  using mutable_T = typename std::remove_reference<T>::type;
  // 3 primitive functions:
  T* begin() const { return b; }
  T* end() const { return e; }
  array_view(T* s, T* f):b(s), e(f) {};
  // everything else based on them:
  size_t size() const { return end()-begin(); }
  array_view(T* s, size_t l):array_view(s,s+l) {}
  array_view():array_view(nullptr,  nullptr) {}
  // repeat next 3 for string, array, initializer list, C array as required:
  template<class A>
  array_view( std::vector<T,A>& v ):array_view(v.data(), v.size()) {}
  // may not compile for non-const T, but that is ok  you get an error:
  template<class A>
  array_view( std::vector<mutable_T,A>const & v ):array_view(v.data(), v.size()) {}
  // in a better world, you'd SFINAE remove the above from consideration
  // consider it for your next iteration of array_view.
  // conversion to containers:
  template<class A>
  explicit operator std::vector<mutable_T,A>() const {
    return convert_to< std::vector<mutable_T, A> >();
  }
  template<class C>
  C convert_to() const {
    C retval(begin(), end());
    return retval;
  }

  // utility functions:
  T& front() const { return *begin(); }
  T& back() const { return std::prev(*end()); }
  // Also rbegin, rend, and whatever else you find you are missing

  // inspired by std::experimental:
  void pop_front() { *this = {std::next(begin()), end()}; }
  void pop_back() { *this = {begin(), std::prev(end())}; }
  // but I find direct use of `view = {x,y}` almost as good
  // these kind of operations are the ONLY ones that are non-const
  // remember this is a view.  If you want a view of const data, make it
  // an array_view<const T>, not a const array_view<T>.  It really works
  // out better.
private:
  T* b
  T* e;
};

the above sample code is not tested.



来源:https://stackoverflow.com/questions/29007753/pass-a-string-recursively-without-recreation

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