问题描述
作为我的厕所的一部分，在C ++标准ANSI ISO IEC 14882 2003，我发现了以下：
As part of my toilet reading on the C++ Standard ANSI ISO IEC 14882 2003, I came across the following:
当我得到一个本地类型和一个复合类型，什么是未命名类型？如果一个类型未命名，你怎么可能试图在模板中使用它，这促使标准口头排除它？
While I get what a local type and a compound type are, what is an unnamed type? If a type is unnamed, how could you even attempt to use it in a template anyway, which prompted the standard to verbally exclude it?
推荐答案
未命名类型真的意味着未命名的枚举或类类型[有关更多信息，请参阅对此回答的评论]。枚举或类类型不必具有名称。例如：
"Unnamed type" really means "unnamed enumeration or class type" [for more information, see the comments to this answer]. An enumeration or class type doesn't have to have a name. For example:
struct { int i; } x; // x is of a type with no name
您可以尝试使用未命名的类型作为模板argument through argument deduction：
You could try to use an unnamed type as a template argument through argument deduction:
template <typename T> void f(T) { }
struct { int i; } x;
f(x); // would call f<[unnamed-type]>() and is invalid in C++03
请注意，此限制已在C ++ 0x中解除，因此此将有效（您也可以使用本地类型作为类型模板参数）。在C ++ 0x中，您还可以使用 decltype 来命名未命名的类型：
Note that this restriction has been lifted in C++0x, so this will be valid (you'll also be able to use local types as type template parameters). In C++0x, you could also use decltype to "name" an unnamed type:
template <typename T> void g() { }
struct { int i; } x;
f<decltype(x)>(); // valid in C++0x (decltype doesn't exist in C++03)
这篇关于什么是C ++中的未命名类型？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！