No member named 'begin' in namespace '

2020-04-08 12:31发布

问题:

I successfully compiled on Windows a code that should be crossplatform. Now when compiling it in Xcode with Mac OS X, I get:

std::valarray<float> v(32);
...
std::sort(begin(v), end(v));            # Use of undeclared identifier 'begin'
std::sort(std::begin(v), std::end(v));  # No member named 'begin' in namespace 'std'
std::sort(std::valarray::begin(v), std::valarray::end(v));  # Idem, error as well

Why does the error No member named 'begin' in namespace 'std' happen?

回答1:

std::begin was introduced with C++11. See this answer for how to enable C++11 in XCode 4.2 (the precise name of the dialect has probably changed by now).

Alternatively, if you can't move to C++11, switch to std::vector and use v.begin().



回答2:

You could be compiling in C++03 mode. Work out how to get your IDE to compile in C++11 mode. XCode 4.2 turn on C++11 may help.

std::sort(std::valarray::begin(v), std::valarray::end(v)); -- I don't think the standard demands this ever work. I guess if valarray implemented begin and end as statics or Koenig friend operators or somesuch.

std::valarray doesn't come with member begin/end. The only way in C++03 to iterate over it was to use [] or with pointers.

valarray is guaranteed to be contiguous.. So you can write

namespace notstd {
  // addressof taken from http://en.cppreference.com/w/cpp/memory/addressof
  template<class T>
  T* addressof(T& arg) {
    return reinterpret_cast<T*>(
           &const_cast<char&>(
              reinterpret_cast<const volatile char&>(arg)));
  }

  template<class T>
  T* begin( std::valarray<T>& v ) { return addressof(v[0]); }
  template<class T>
  T* end( std::valarray<T>& v ) { return begin(v)+v.size(); }
  template<class T>
  T const* begin( std::valarray<T> const& v ) { return addressof(v[0]); }
  template<class T>
  T const* end( std::valarray<T> const& v ) { return begin(v)+v.size(); }
}

and use notstd::begin on your valarray.