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 我问你是否可以做一些事情： 
 
 
Foo<decltype([]()->void { })> foo;
 
 不，你不能，因为lambda表达式不会出现在未评估的上下文中（例如decltype和sizeof等）。 C ++ 0x FDIS，5.1.2 [expr.prim.lambda] p2 
 
 
 
 lambda表达式的评估导致临时值（12.2）。 这个临时被称为闭包对象。 lambda表达式不得出现在未经评估的操作数 （第5条）中。 [注意：闭包对象的行为就像一个函数对象（20.8）.- end note] （强调我的） 
 
 
 您需要首先创建一个特定的lambda，然后使用decltype： 
 
auto my_comp = [](const std::string& left, const std::string& right) -> bool {
  // whatever
}

typedef std::unordered_map<
  std::string,
  std::string,
  std::hash<std::string>,
  decltype(my_comp)
  > map_type;
 
 那是因为每个lambda派生的闭包对象可能有完全不同的类型，毕竟它们就像匿名函数。 

 
 
I'm asking if you can do something like:
 
 
Foo<decltype([]()->void { })> foo;
 
No you can't, because lambda expressions shall not appear in an unevaluated context (such as decltype and sizeof, amongst others). C++0x FDIS, 5.1.2 [expr.prim.lambda] p2
 
 
 
The evaluation of a lambda-expression results in a prvalue temporary (12.2). This temporary is called the closure object. A lambda-expression shall not appear in an unevaluated operand (Clause 5). [ Note: A closure object behaves like a function object (20.8).—end note ] (emphasis mine)
 
 
You would need to first create a specific lambda and then use decltype on that:
 
auto my_comp = [](const std::string& left, const std::string& right) -> bool {
  // whatever
}

typedef std::unordered_map<
  std::string,
  std::string,
  std::hash<std::string>,
  decltype(my_comp)
  > map_type;
 
That is because each lambda-derived closure object could have a completely different type, they're like anonymous functions after all.