如何使用emplace（）在std :: map的值是std :: set（地图从一个集合）？ [英] How to use emplace() in a std::map whose value is a std::set (map from something to a set)?
问题描述
我有一个 std :: map< int，std :: set< int>> 命名为 misi 。我想知道为什么 misi.emplace（2345，{6,9}）; 和 misi.emplace（{2345，{6，9} }
I have a std::map<int, std::set<int>> named misi. I'm wondering why misi.emplace(2345, {6, 9}); and misi.emplace({2345, {6, 9}}); don't work as expected, as shown below.
#include <set> // std:set
#include <map> // std::map
#include <utility> // std::piecewise_construct, std::pair
#include <tuple> // std::forward_as_tuple
#include <iostream> // std::cout, std::endl
int main()
{
// --- std::set initializer list constructor ---
std::set<int> si({42, 16});
std::cout << "si.size(): " << si.size() << std::endl; // 2
std::cout << "*si.begin(): " << *si.begin() << std::endl; // 16
// --- std::set emplace() ---
si.emplace(7);
std::cout << "si.size(): " << si.size() << std::endl; // 3
std::cout << "*si.begin(): " << *si.begin() << std::endl; // 7
std::cout << "--------" << std::endl;
以上是 std :: set 可以看到初始化列表构造函数和 emplace（）工作完美。
Above is std::set, you can see that the initializer list constructor and emplace() work perfectly.
// --- std::map initializer list constructor ---
std::map<int, int> mii({ {0, 42}, {1, 16} });
std::cout << "mii.size(): " << mii.size() << std::endl; // 2
std::cout << "mii[0]: " << mii[0] << std::endl; // 42
std::cout << "mii[1]: " << mii[1] << std::endl; // 16
// --- std::map emplace() ---
mii.emplace(1234, 7);
std::cout << "mii.size(): " << mii.size() << std::endl; // 3
std::cout << "mii[1234]: " << mii[1234] << std::endl; // 7
// --- std::map emplace() with std::pair() ---
mii.emplace(std::pair<int, int>(2345, 6));
std::cout << "mii.size(): " << mii.size() << std::endl; // 4
std::cout << "mii[2345]: " << mii[2345] << std::endl; // 6
std::cout << "--------" << std::endl;
以上是 std :: map of int 到 int ，你可以看到方法工作完美，除了在最后一个例子中， std :: pair 可能有点冗余。我不知道 std :: pair 是否在原地构造。 （我想不是）
Above is std::map of int to int, you can see the methods work perfectly too, except that in the last example, the std::pair might be somehow redundant. I wonder if the std::pair is constructed in-place or not. (Well, I guess not)
// --- std::map to std::set initializer list constructor ---
std::map<int, std::set<int>> misi({ {0, {42, 16}}, {1, {7}} });
std::cout << "misi.size(): " << misi.size() << std::endl; // 2
std::cout << "*misi[0].begin(): " << *misi[0].begin() << std::endl; // 16
std::cout << "*misi[1].begin(): " << *misi[1].begin() << std::endl; // 7
对于 std :: map 到 std :: set ，初始化列表构造函数完美地工作，如上所示。但是 emplace（）不会！ （如下所示）
For a std::map to a std::set, the initializer list constructor works perfectly as shown above. But the emplace() does not! (as shown below)
// --- Compilation Errors ---
//misi.emplace(2345, 6, 9);
//misi.emplace({2345, 6, 9});
//misi.emplace(2345, {6, 9});
//misi.emplace({2345, {6, 9}});
//misi.emplace(
// std::piecewise_construct,
// std::forward_as_tuple(2345),
// std::forward_as_tuple(6, 9)
//);
//misi.emplace(
// std::piecewise_construct,
// std::forward_as_tuple(2345),
// std::forward_as_tuple({6, 9})
//);
这里，以下语法是好的，但是不能完全按照我想要的方式：
Here, the following syntax is okay, but doesn't do exactly what I wanted:
// --- OK ---
misi.emplace(std::pair<int, std::set<int>>(2345, {6, 9}));
std::cout << "misi.size(): " << misi.size() << std::endl; // 3
std::cout << "*misi[2345].begin(): " << *misi[2345].begin() << std::endl; //6
std::cout << "--------" << std::endl;
return 0;
}
所以，似乎没有办法创建 std :: pair 就地，似乎 std :: set 是就地创建的（右？）。有人有任何想法吗？
So, it seems that there is no way to create the std::pair in-place, and it seems that the std::set is created in-place (right?). Does anyone have any ideas?
我使用的编译器是：
$ clang++ --version
Apple LLVM version 6.0 (clang-600.0.56) (based on LLVM 3.5svn)
Target: x86_64-apple-darwin13.4.0
Thread model: posix
推荐答案
支持的初始设置列表没有类型，不能完全转发。在这种特殊情况下，如果你想要构建所有东西，你可以明确指定类型（ std :: initializer_list< int> ）：
Braced initializer lists have no type, so they can't be perfectly forwarded. In this particular case, you can specify the type (std::initializer_list<int>) explicitly if you want everything to be constructed in place:
misi.emplace(
std::piecewise_construct,
std::forward_as_tuple(2345),
std::forward_as_tuple(std::initializer_list<int>{6, 9})
);
由于只有一个参数传递给键和值，而 initializer_list 构造函数 std :: set 不是 explicit ，可以删除 piecewise_construct 并使用构造函数构造构造对两个参数：
Since only one argument each is passed for the key and the value, and the initializer_list constructor of std::set is not explicit, you can remove the piecewise_construct altogether and have emplace construct the pair using the constructor taking two arguments:
misi.emplace(
2345,
std::initializer_list<int>{6, 9}
);
mii.emplace(std::pair<int, int>(2345, 6));
我不知道std :: pair是否在原地构造。 （Well，I
guess not）
I wonder if the std::pair is constructed in-place or not. (Well, I guess not)
不，临时 std :: pair< int ，int> 被构造并传递到 emplace ，其构造映射它的 value_type （它是 pair< const int，int> ）。后者是实际存储在映射中的内容。
No, a temporary std::pair<int, int> is constructed and passed to emplace, which constructs an instance of the map's value_type (which is pair<const int, int>) with it. The latter is what is actually stored in the map.
misi.emplace(std::pair<int, std::set<int>>(2345, {6, 9}));
似乎std :: set是在原地创建的>
it seems that the std::set is created in-place (right?).
否。再次，这将创建一个临时 std :: pair< int，std :: set< int>> ，然后构造实际存储在 map （它是 pair< const int，std :: set< int>> ）。第二个构造将执行从存储在临时中的 set< int> 的移动。
No. Again, this creates a temporary std::pair<int, std::set<int>>, and then constructs what's actually stored in the map (which is pair<const int, std::set<int>>) with it. This second construction will perform a move from the set<int> stored in the temporary.
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