initializer_list with auto contains multiple expressions

谁都会走 提交于 2019-12-05 09:26:41

问题


Rather simple question,

auto x11 {1,2,3,4};
auto x1 = {1,2,3,4};
auto x22 {1.0, 2.25, 3.5};
auto x2 = {1.0, 2.25, 3.5};

As far as I understand, there should be no difference here with respect to having = or not. However, using llvm/clang 6.0.0 (with --std=c++17), I get :

main1.cpp:35:17: error: initializer for variable 'x11' with type 'auto' contains multiple
  expressions
auto x11 {1,2,3,4};
~~~~~~~~    ^

main1.cpp:37:20: error: initializer for variable 'x22' with type 'auto' contains multiple
  expressions
auto x22 {1.0, 2.25, 3.5};

From Stroustroup's C++ book, page.162:

auto x1 {1,2,3,4}; // x1 is an initializer_list<int>
auto x2 {1.0, 2.25, 3.5 }; // x2 is an initializer_list of<double>

So, is there really a problem in not having = in there?


回答1:


The rule of auto type deduction changed since C++17.

(since C++17)
In direct-list-initialization (but not in copy-list-initalization), when deducing the meaning of the auto from a braced-init-list, the braced-init-list must contain only one element, and the type of auto will be the type of that element:

auto x1 = {3}; // x1 is std::initializer_list<int>
auto x2{1, 2}; // error: not a single element
auto x3{3};    // x3 is int
               // (before C++17 x2 and x3 were both std::initializer_list<int>)

So before C++17, all the variables in your sample work fine and have type std::initializer_list<int>. But since C++17, for direct initialization (i.e. for x11 and x22) the braced-initializer must contain only one element (and their type would be the type of the element) then become ill-formed code.

See N3922 and N3681 for more.



来源:https://stackoverflow.com/questions/51166807/initializer-list-with-auto-contains-multiple-expressions

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!