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我有两个boost::hana::map要加入。

constexpr auto m1 = hana::make_map( 
    hana::make_pair("key1"_s, hana::type_c<std::string>), 
    hana::make_pair("key2"_s, hana::type_c<std::string>) 
); 

constexpr auto m2 = hana::make_map( 
    hana::make_pair("key3"_s, hana::type_c<std::string>), 
    hana::make_pair("key4"_s, hana::type_c<std::string>), 
    hana::make_pair("key5"_s, hana::type_c<std::string>) 
);

如何获取包含这两者的地图，如下所示？

constexpr auto result = hana::make_map( 
    hana::make_pair("key1"_s, hana::type_c<std::string>), 
    hana::make_pair("key2"_s, hana::type_c<std::string>), 
    hana::make_pair("key3"_s, hana::type_c<std::string>), 
    hana::make_pair("key4"_s, hana::type_c<std::string>), 
    hana::make_pair("key5"_s, hana::type_c<std::string>) 
);
4

1 回答 1

2

较新的版本

（Boost.Hana > 1.2，Boost >= 1.65）

有boost::hana::union_加入boost::hana::maps的功能。

返回两个地图的并集。

给定两个映射xs和ys，hana::union_(xs, ys)是一个新映射，其中包含 的所有元素xs和 的所有元素ys，没有重复。如果两者xs和ys都包含具有相同键的元素，则采用其中的一个ys。

引用自 Boost.Hana 文档

还有例子：

// Simple example
constexpr auto m1 = hana::make_map(
    hana::make_pair("key1"_s, hana::type_c<std::string>),
    hana::make_pair("key2"_s, hana::type_c<std::string>)
);
constexpr auto m2 = hana::make_map(
    hana::make_pair("key3"_s, hana::type_c<std::string>),
    hana::make_pair("key4"_s, hana::type_c<std::string>),
    hana::make_pair("key5"_s, hana::type_c<std::string>)
);
BOOST_HANA_CONSTANT_CHECK(hana::union_(m1, m2) == hana::make_map(
       hana::make_pair("key1"_s, hana::type_c<std::string>),
       hana::make_pair("key2"_s, hana::type_c<std::string>),
       hana::make_pair("key3"_s, hana::type_c<std::string>),
       hana::make_pair("key4"_s, hana::type_c<std::string>),
       hana::make_pair("key5"_s, hana::type_c<std::string>)
));

// Example with overwriting pair (the same key)
constexpr auto m3 = hana::make_map(
    hana::make_pair(hana::type_c<int>, hana::int_c<1>),
    hana::make_pair(hana::type_c<bool>, hana::bool_c<true>)
);
constexpr auto m4 = hana::make_map(
       hana::make_pair(hana::type_c<char>, hana::char_c<'c'>),
       hana::make_pair(hana::type_c<bool>, hana::bool_c<false>)
);
BOOST_HANA_CONSTANT_CHECK(hana::union_(m3, m4) == hana::make_map(
       hana::make_pair(hana::type_c<int>, hana::int_c<1>),
       hana::make_pair(hana::type_c<bool>, hana::bool_c<false>),
       hana::make_pair(hana::type_c<char>, hana::char_c<'c'>)
));

Godbolt在线编译器上的完整工作交互示例。

旧版本

（Boost.Hana <= 1.1，Boost <= 1.64）

基于一些 Boost 论坛主题的修改示例。

constexpr auto map1 = hana::make_map( 
    hana::make_pair(hana::type_c<TwoOtherKeys1>,      hana::type_c<SomeValue>),
    hana::make_pair(hana::type_c<SameKey_SameType>,   hana::type_c<SomeValue>),
    hana::make_pair(hana::type_c<SameKey_DifferType>, hana::type_c<SomeValue>)
);
constexpr auto map2 = hana::make_map( 
    hana::make_pair(hana::type_c<TwoOtherKeys2>,      hana::type_c<SomeValue>), 
    hana::make_pair(hana::type_c<SameKey_SameType>,   hana::type_c<SomeValue>), 
    hana::make_pair(hana::type_c<SameKey_DifferType>, hana::type_c<OtherValue>) 
); 

// Creates result sequence by inserting all of (source) `map1` sequence into (initial) `map1` sequence. 
// Since both sequences are maps with unique keys, keys that exist in both source and initial map got overwrited.
constexpr auto result = hana::fold(map1, map2, hana::insert);

// Excepted result
constexpr auto expected = hana::make_map(
    // Diffrent keys pairs are copied.
    hana::make_pair(hana::type_c<TwoOtherKeys1>, hana::type_c<SomeValue>),
    hana::make_pair(hana::type_c<TwoOtherKeys2>, hana::type_c<SomeValue>),

    // The same key and type pairs are just passed.
    hana::make_pair(hana::type_c<SameKey_SameType>, hana::type_c<SomeValue>),   

    // The same key, differ type - overwrited.
    hana::make_pair(hana::type_c<SameKey_DifferType>, hana::type_c<OtherValue>)
);

// Assertion to confirm exceptation
static_assert(result == expected, ""); // OK

Godbolt在线编译器上的完整工作交互示例。

该解决方案与最新版本的 Boost.Hana 中的解决方案完全相同。

全部归功于Louis Dionne的解决方案。

于 2019-02-13T02:28:34.200 回答