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      2. 通用 lambda 的熟悉模板语法

        Familiar template syntax for generic lambdas(通用 lambda 的熟悉模板语法)

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                  本文介绍了通用 lambda 的熟悉模板语法的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

                  问题描述

                  对于 c++20,建议为通用 lambdas 添加以下语法 p0428r2.pdf

                  For c++20 it is proposed to add the following syntax for generic lambdas p0428r2.pdf

                  auto f = []<typename T>( T t ) {};
                  

                  但是当前在 gcc 8 中的实现不接受以下实例化:

                  But the current implementation in gcc 8 did not accept the following instantiation:

                  f<std::string>("");
                  

                  这是 gcc 中的实现错误还是缺少语言功能?我知道我们谈论的是提案而不是批准的规范.

                  Is that a implementation bug in gcc or a missing language feature? I know we talk about a proposal and not a approved specification.

                  完整示例(与模板函数语法比较):

                  Complete example ( with comparison to template function syntax ):

                  template <typename T> void n( T t ) { std::cout << t << std::endl; }
                  
                  auto f = []<typename T>( T t ) { std::cout << t << std::endl; };
                  
                  int main()
                  {
                      f<std::string>("Hello");  // error!
                      n<std::string>("World");
                  }
                  

                  抱怨以下错误:

                  main.cpp:25:22: 错误:'>' 标记前的预期主表达式f("你好");

                  main.cpp:25:22: error: expected primary-expression before '>' token f("Hello");

                  推荐答案

                  lambda 表达式的结果不是函数;它是一个函数对象.也就是说,它是一个具有 operator() 重载的类类型.所以这个:

                  The result of a lambda expression is not a function; it is a function object. That is, it is a class type that has an operator() overload on it. So this:

                  auto f = []<typename T>( T t ) {};
                  

                  相当于:

                  struct unnamed
                  {
                    template<typename T>
                    void operator()(T t) {}
                  };
                  
                  auto f = unnamed{};
                  

                  如果你想显式地为 lambda 函数提供模板参数,你必须显式调用 operator():f.operator()