c++ - 通用 lambda 在 C++14 中如何工作？

Question

通用 lambda 在 C++14 标准中如何工作(auto 关键字作为参数类型)？

它是基于 C++ 模板，对于每个不同的参数类型，编译器生成一个具有相同主体但替换类型(编译时多态性)的新函数，还是更类似于 Java 的泛型(类型删除)？

代码示例:

auto glambda = [](auto a) { return a; };

Answer 1

通用 lambda 是在 C++14 中引入的。

简单地说，由 lambda 表达式定义的闭包类型将具有一个模板化 调用运算符，而不是 C++11 的常规非模板调用运算符> 的 lambda(当然，当 auto 在参数列表中至少出现一次时)。

所以你的例子:

auto glambda = [] (auto a) { return a; };

将使 glambda 成为这种类型的实例:

class /* unnamed */
{
public:
    template<typename T>
    T operator () (T a) const { return a; }
};

C++14 标准草案 n3690 第 5.1.2/5 段规定了给定 lambda 表达式的闭包类型的调用运算符是如何定义的:

The closure type for a non-generic lambda-expression has a public inline function call operator (13.5.4) whose parameters and return type are described by the lambda-expression’s parameter-declaration-clause and trailing-return-type respectively. For a generic lambda, the closure type has a public inline function call operator member template (14.5.2) whose template-parameter-list consists of one invented type template-parameter for each occurrence of auto in the lambda’s parameter-declaration-clause, in order of appearance. The invented type template-parameter is a parameter pack if the corresponding parameter-declaration declares a function parameter pack (8.3.5). The return type and function parameters of the function call operator template are derived from the lambda-expression’s trailing-return-type and parameter-declarationclause by replacing each occurrence of auto in the decl-specifiers of the parameter-declaration-clause with the name of the corresponding invented template-parameter.

最后:

Is it similar to templates where for each different argument type compiler generates functions with the same body but changed types or is it more similar to Java's generics?

正如上一段所解释的，通用 lambda 只是具有模板化调用运算符的独特、未命名仿函数的语法糖。那应该回答你的问题:)