问题描述
考虑功能:
template<typename T>
void printme(T&& t) {
for (auto i : t)
std::cout << i;
}
或任何其他需要一个参数为 begin()/end() - 启用类型的函数.
or any other function that expects one parameter with a begin()/end() - enabled type.
为什么以下内容是非法的?
printme({'a', 'b', 'c'});
当所有这些都合法时:
printme(std::vector<char>({'a', 'b', 'c'}));
printme(std::string("abc"));
printme(std::array<char, 3> {'a', 'b', 'c'});
我们甚至可以这样写:
const auto il = {'a', 'b', 'c'};
printme(il);
或
printme<std::initializer_list<char>>({'a', 'b', 'c'});
推荐答案
你的第一行 printme({'a', 'b', 'c'}) 是非法的，因为模板参数 T 无法推断.如果您明确指定模板参数，它将起作用，例如printme<vector>({'a', 'b', 'c'})</vector 或 printme<initializer_list>({'a', 'b'', 'c'})</initializer_list.
Your first line printme({'a', 'b', 'c'}) is illegal because the template argument T could not be inferred. If you explicitly specify the template argument it will work, e.g. printme<vector<char>>({'a', 'b', 'c'}) or printme<initializer_list<char>>({'a', 'b', 'c'}).
您列出的其他参数是合法的，因为参数具有明确定义的类型，因此模板参数 T 可以很好地推导出来.
The other ones you listed are legal because the argument has a well-defined type, so the template argument T can be deduced just fine.
带有 auto 的代码段也有效，因为 il 被认为是 std::initializer_list<char> 类型，因此模板可以推导出 printme() 的参数.
Your snippet with auto also works because il is considered to be of type std::initializer_list<char>, and therefore the template argument to printme() can be deduced.
唯一的搞笑"这里的部分是 auto 将选择类型 std::initializer_list 但模板参数不会.这是因为 C++11 标准的 § 14.8.2.5/5 明确指出这是模板参数的非推导上下文:
The only "funny" part here is that auto will pick the type std::initializer_list<char> but the template argument will not. This is because § 14.8.2.5/5 of the C++11 standard explicitly states that this is a non-deduced context for a template argument:
一个函数参数，其关联参数是一个初始化列表 (8.5.4)，但该参数没有 std::initializer_list 或对可能具有 cv 限定的 std::initializer_list 类型的引用.[示例:
A function parameter for which the associated argument is an initializer list (8.5.4) but the parameter does not have std::initializer_list or reference to possibly cv-qualified std::initializer_list type. [Example:
template<class T> void g(T);
g({1,2,3}); // error: no argument deduced for T
——结束示例 ]
但是对于 auto，第 7.1.6.4/6 节明确支持 std::initializer_list<>
However with auto, § 7.1.6.4/6 has explicit support for std::initializer_list<>
如果初始化器是一个 braced-init-list (8.5.4)，带有 std::initializer_list.
if the initializer is a braced-init-list (8.5.4), with
std::initializer_list<U>.
这篇关于initializer_list 和模板类型推导的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，WP2