initializer_list with auto contains multiple expre

2019-04-03 05:35发布

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?

1条回答
兄弟一词,经得起流年.
2楼-- · 2019-04-03 05:49

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.

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