C ++ 11允许向量< const T>？ [英] Does C++11 allow vector<const T>?

问题描述

容器需求已从C ++ 03更改为C ++ 11。虽然C ++ 03有一些要求（例如拷贝构造性和向量的可分配性），但C ++ 11定义了每个容器操作的细粒度要求（第23.2节）。

Container requirements have changed from C++03 to C++11. While C++03 had blanket requirements (e.g. copy constructibility and assignability for vector), C++11 defines fine-grained requirements on each container operation (section 23.2).

结果，你可以eg在向量中存储一个可复制构造但不可分配的类型（例如具有const成员的结构），只要执行某些不需要赋值的操作（构造和 push_back 是这样的操作; insert 不是）。

As a result, you can e.g. store a type that is copy-constructible but not assignable - such as a structure with a const member - in a vector as long as you only perform certain operations that do not require assignment (construction and push_back are such operations; insert is not).

我想知道的是：意味着标准现在允许向量< const T> ？我没有看到任何理由，它不应该 - const T ，就像一个结构与一个const成员，是一个类型是复制构造，但不可分配 - 但我可能错过了一些东西。

What I'm wondering is: does this mean the standard now allows vector<const T>? I don't see any reason it shouldn't - const T, just like a structure with a const member, is a type that is copy constructible but not assignable - but I may have missed something.

（部分原因是让我觉得我可能错过了一些东西，是gcc中断崩溃和烧伤，如果你尝试实例化 vector< const T> ，但是对于向量< T> 很好，其中T有一个const成员）。

(Part of what makes me think I may have missed something, is that gcc trunk crashes and burns if you try to instantiate vector<const T>, but it's fine with vector<T> where T has a const member).

推荐答案

不，我相信分配器要求说T可以是非常量，非引用对象类型。

No, I believe the allocator requirements say that T can be a "non-const, non-reference object type".

你不能用常量对象的向量做很多事情。并且 const向量< T> 几乎是相同的。

You wouldn't be able to do much with a vector of constant objects. And a const vector<T> would be almost the same anyway.

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