非本地Lambda和捕捉变量 - 什么是“块范围”是指

Question

我目前与C++ 11个lambda表达式玩，我发现我无法理解一个例子。根据标准：

A lambda-expression whose smallest enclosing scope is a block scope (3.3.3) is a local lambda expression; any other lambda-expression shall not have a capture-list in its lambda-introducer

所以，我创建了简单的例子：http://ideone.com/t9emu5

我期待这个代码将不会因为在拍摄变量的编译：

int a = 10; 
auto x = [a] { return 1;}; 
int main() { 
    int k = 5; 
    auto p = [k]{ return k; }; 
    return 0; 
} 

在ideone代码非块范围（或者至少认为auto x = ...部分不在块范围内）。但代码正在编译 - 可以吗？

如果是正常 - 块范围是什么？

（我不知道我所用的编译器版本，因为目前我只有ideone站点的访问。

感谢解释！

Answer 1

看起来这是一个编译器扩展。G ++ 4.8 0.1编译此同时发出警告：

warning: capture of variable ‘a’ with non-automatic storage duration [enabled by default]

铛++ 3.4不能编译如下：

error: 'a' cannot be captured because it does not have automatic storage duration

均指[expr.prim.lambda]/10

The identifiers in a capture-list are looked up using the usual rules for unqualified name lookup (3.4.1); each such lookup shall find a variable with automatic storage duration declared in the reaching scope of the local lambda expression.

似乎他们不额外检查拉姆达的封闭范围内，我可以想像这将是多余的（没有的变量名自动存储持续时间在非块/命名空间范围）。

---

块范围在[basic.scope.block]/1

A name declared in a block (6.3) is local to that block; it has block scope.

定义和块定义为：

So that several statements can be used where one is expected, the compound statement (also, and equivalently, called “block”) is provided.

     compound-statement:
         {statement-seq_opt}

所以，你说得对，你的在块范围内全局声明的lambda是而不是。