c++ - template argument deduction - boost::unordered_map -


when reading boost::unordered_map source code, i'was confused definition of class, below:

template <class k, class t, class h, class p, class a> class unordered_map {   public:     typedef k key_type;     typedef std::pair<const k, t> value_type;     typedef t mapped_type;     typedef h hasher;     typedef p key_equal;     typedef allocator_type; private:     typedef boost::unordered::detail::map<a, k, t, h, p> types;     typedef typename types::traits allocator_traits;     typedef typename types::table table; ... private:     table table_;  public:     // constructors     explicit unordered_map(             size_type = boost::unordered::detail::default_bucket_count,             const hasher& = hasher(),             const key_equal& = key_equal(),             const allocator_type& = allocator_type()); ... };

there's no default value class h, class p , class a, why user can declare map instance boost::unordered_map<key, value> map? didn't find guide. help? detailed document/link best.

if start looking @ first header file, 1 #include, going find declaration template, specifies defaults template parameters. (that declaration may not in actual header file, in internal header file being #included there).

the declaration gets compiled first, , declares defaults, , you're seeing here actual template definition.

the initial template declaration specifies defaults template parameters, , not need specified in template definition (and, in fact, compilation error if did).

a brief example:

// initial declaration  template<typename t=int> class foo;  // definition  template<typename t> class foo {};  // usage foo<> bar;

the actual boost header files quite complex in structure, declaration , definition of templates spread across separate files, various sundry reasons, assembled in right order actual header file #include in code.


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