英文:
C++ Inferring Template Arguments of Type Inside of Angle Brackets
问题
这行代码中的类型注解是否有简化的方法：
std::array<std::tuple<char, int>, 3> arr {{{'a', 1}, {'b', 2}, {'c', 3}}};
我认为这样写应该可以：
std::array<std::tuple> arr {{{'a', 1}, {'b', 2}, {'c', 3}}};
但实际上不行。
英文:
Is there anyway to simplify the type annotation in this line of code:
std::array<std::tuple<char, int>, 3> arr {{{'a', 1}, {'b', 2}, {'c', 3}}};
I thought this would work:
std::array<std::tuple> arr {{{'a', 1}, {'b', 2}, {'c', 3}}};
But it does not.
答案1
得分: 2
A placeholder for a deduced class type, i.e. the name of a class template without a template argument list like std::tuple, cannot be used as a template argument.
CTAD (class template argument deduction) is only possible if the whole specified type is a placeholder for a deduced class type, e.g.
std::array{/*...*/}
where std::array is the placeholder naming a class template without template argument list. The template argument list must not be present at all. Even std::array<int>{0, 1, 2, 3} isn't possible.
The syntax you want isn't possible. You need to make some concessions like using std::make_tuple around each element of the array initializer.
英文:
A placeholder for a deduced class type, i.e. the name of a class template without a template argument list like std::tuple, cannot be used as a template argument.
CTAD (class template argument deduction) is only possible if the whole specified type is a placeholder for a deduced class type, e.g.
std::array{/*...*/}
where std::array is the placeholder naming a class template without template argument list. The template argument list must not be present at all. Even std::array<int>{0, 1, 2, 3} isn't possible.
The syntax you want isn't possible. You need to make some concessions like using std::make_tuple around each element of the array initializer.
答案2
得分: 2
你可以使用 std::to_array 来简化为
auto arr = std::to_array({std::tuple{'a', 1}, {'b', 2}, {'c', 3}});
英文:
You can use std::to_array to simplify to
auto arr = std::to_array({std::tuple{'a', 1}, {'b', 2}, {'c', 3}});
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