decltype 的行为

Behavior of decltype(decltype 的行为)

问题描述

假设我有一些 stl 容器类 obj 的对象.我可以通过这种方式定义其他相同类型的对象:

decltype(obj) obj2;
但我不能以这种方式为容器声明迭代器:
decltype(obj)::iterator it = obj.begin();
为什么?我做错了什么吗?
 解决方案 
根据最终的 C++0x 草案 (FDIS)，您的代码格式良好.这是 Visual Studio 编译器尚未实现的后期更改.
与此同时，一种解决方法是使用 typedef:
typedef decltype(obj) obj_type;obj_type::iterator it = obj.begin();
相关章节和诗句是5.1.1/8:
<前>合格 ID:[...]nested-name-specifier 模板opt unqualified-id嵌套名称说明符:[...]声明类型说明符 ::声明类型说明符:decltype ( 表达式 )

为了完整起见:

原始核心问题

措辞建议

Say I have an object of some of stl container classes obj. I can define other object of same type this way:

decltype(obj) obj2;

But I can't declare iterator for the container this way:

decltype(obj)::iterator it = obj.begin();

Why? Am I doing something wrong?

Your code is well-formed according to the final C++0x draft (FDIS). This was a late change that's not yet been implemented by the Visual Studio compiler.

In the meantime, a workaround is to use a typedef:

typedef decltype(obj) obj_type;
obj_type::iterator it = obj.begin();

EDIT: The relevant chapter and verse is 5.1.1/8:

qualified-id:
    [...]
    nested-name-specifier templateopt unqualified-id

nested-name-specifier:
    [...]
    decltype-specifier ::

decltype-specifier:
    decltype ( expression )

And for completeness's sake:

The original core issue

Proposal for wording

这篇关于decltype 的行为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，也希望大家多多支持编程学习网！