Exercise 4-5

Return different types based on parameter types. More...

Files
file	exercise-4-5.hpp
	Solution to Exercise 4-5.

Namespaces
namespace	exercise_4_5
	Encapsulate solution for Exercise 4-5.

Functions
static void	anonymous_namespace{chapter-4.cpp}::test_exercise_4_5 ()
	Tests for Exercise 4-5.

Detailed Description

Return different types based on parameter types.

4-5. Consider the following function template, which is designed to provide a
     "container-based" (as opposed to iterator-based) interface to std::find:
         template <typename Container, typename Value>
         typename Container::iterator
         container_find(Container& c, Value const& v)
         {
             return std::find(c.begin(), c.end(), v);
         }
     As coded, container_find won't work for const containers; Container will
     be deduced as const X for some container type X, but when we try to
     convert the Container::const_iterator returned by std::find into a
     Container::iterator, compilation will fail.  Fix the problem using a
     small metaprogram to compute container_find's return type.

Function Documentation

test_exercise_4_5()

static void anonymous_namespace{chapter-4.cpp}::test_exercise_4_5 ( )

static

Tests for Exercise 4-5.

Definition at line 164 of file chapter-4.cpp.