问一下Lambda表达式

[karorokiki]

Lambda表达式 实际是类么 和函数对象 一样么

[healer_kx]

我还没好好研究呢，我感觉可以理解为一个函数对象？

[a707000646]

http://baike.baidu.com/view/3580897.htm

[taodm]

楼主散分结束，可以结贴了。

[iamnobody]

匿名的函数对象，楼上正解

[sanae]

template <typename T>
void foo(T bar){
bar();
}

auto x = [](){...};
foo(x);

gcc4.6.1中 实例化的T和推断auto都是一个类,它包含了刚才写的lambda作为operator (),
lambda也就是operator ()是（非静态）成员函数，this指针似乎与capture的内容有关……

[FrankHB1989]

是。
ISO C++11
5.1.2
2 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 ]
3 The type of the lambda-expression (which is also the type of the closure object) is a unique, unnamed nonunion class type — called the closure type — whose properties are described below. This class type is not an aggregate (8.5.1). The closure type is declared in the smallest block scope, class scope, or namespace scope that contains the corresponding lambda-expression. [ Note: This determines the set of namespaces and classes associated with the closure type (3.4.2). The parameter types of a lambda-declarator do not affect these associated namespaces and classes. —end note ] An implementation may define the closure type differently from what is described below provided this does not alter the observable behavior of the program other than by changing:
— the size and/or alignment of the closure type,
— whether the closure type is trivially copyable (Clause 9),
— whether the closure type is a standard-layout class (Clause 9), or
— whether the closure type is a POD class (Clause 9).