问题描述
假设我们有一个具有以下接口的对象:
suppose we have an object with the following interface:
struct Node_t {
... const std::vector< something >& getChilds() const;
} node;
现在，我使用 auto 变量访问属性，如下所示:
Now, i access the property with an auto variable like this:
auto childs = node->getChilds();
childs 的类型是什么?一个 std::vector 或对一个的引用?
what is the type of childs? a std::vector< something > or a reference to one?
推荐答案
childs 的类型将为 std::vector.
auto 由与 模板类型推导相同的规则提供支持.此处选择的类型与为 template 选择的类型相同.f(T t); 在类似 f(node->getChilds()) 的调用中.
auto is powered by the same rules as template type deduction. The type picked here is the same that would get picked for template <typename T> f(T t); in a call like f(node->getChilds()).
类似地，auto& 会为您提供与 template 选择的类型相同的类型.f(T& t); 和 auto&& 将获得与 template 选择的相同类型.f(T&&t);.
Similarly, auto& would get you the same type that would get picked by template <typename T> f(T& t);, and auto&& would get you the same type that would get picked by template <typename T> f(T&& t);.
这同样适用于所有其他组合，例如 auto const& 或 auto*.
The same applies for all other combinations, like auto const& or auto*.
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