c++14 SFINAE dereference iterator

c++14 SFINAE dereference iterator

Original problem: write function f (r), if R is an iterator, then return F (*r), otherwise return R.

summary: questions:

• What is an iterator?

  • iterators are a concept in C + +, and it is an iterator type if type it meets the following conditions

    • can be copied construction ( Copyconstructible)

    • can copy assignment ( Copyassignable)

    • can be destructor ( Destructibale)

    • left value It Object exchangeable (swappable)

    • Std::iterator_traits<it> includes the following types of Members:value_type, difference_type, reference, pointer and ITERATOR_ Category

    • for An lvalue R for It, with the following expression being legal and having the specified meaning:

      • *r return value type: Unspecified Preconditions: r

      • ++r < Span style= "font-size:small" > return value type: It & preconditions: r self-increment

in subsequent implementations, the requirements for iterators are relaxed: R with Lvalue R, or if *r is legal, then there is an iterator type. Meaning That:

write the function F (r), if the Lvalue R can be dereferenced, return F (*r), otherwise return R.

problem Analysis:

The return value type of f changes with the actual parameter. If R is an int * * type, A minimum of two F overloads of int *f (int * *) and int f (int *) are required. These function overloads need to be generated by a call to f at compile time through generics .

when called, the overload resolution stage requires the compiler to select the appropriate function Template.

• t The lvalue can be dereferenced, Select the function template ( 1

   template <class  t>auto f (T r) {
   Span style= "color: #0000ff" >return  f (*r);}  

• Otherwise select a function template ( 2)

   template <class  t>auto f (T r) { 
  return   

question:

• < Span lang= "zh-cn" > How do I select a function template based on the above rules?

  • f templates are valid, Select the template ( 1 Span style= "font-family:noto Sans CJK SC Regular" >

  • legally defined as: r can be dereferenced.

questions:

• How do I express a priority selection ?

  • by overloading an implicit conversion sequence Hierarchy. Sothat the function generated by the template (1) is graded higher than the template (2)

• < Span lang= "zh-cn" > How to express legitimacy?

  • sfinae r< Span style= "font-family:noto Sans CJK SC Regular" > cannot be dereferenced, making the template ( 1 (substitution failure)

implementation:

1#include <cassert>2 3#include <type_traits>4#include <utility>5 6Template <classT>7 Auto F (T x, ...) {8   returnx;9 }Ten  oneTemplate <classTclass= Decltype (*std::d eclval<t> ()) > aAuto F (T x,int) { -   returnF (*x, 0); - } the  - intmain () { -   intx =3, *p = &x; -Assert (f (&p,0)==3); +    -   return 0; +}

text:

Withheld

FAQ:

Not Yet.

references:

Http://en.cppreference.com/w/cpp/concept/Iterator

Http://en.cppreference.com/w/cpp/language/overload_resolution (return Type Deduction)

Http://en.cppreference.com/w/cpp/utility/declval

Http://en.cppreference.com/w/cpp/language/overload_resolution

Http://en.cppreference.com/w/cpp/language/template_argument_deduction

c++14 SFINAE dereference iterator