我们可以使用 lambda 表达式作为函数参数的默认值吗?

2021-12-23 00:00:00 lambda c++ c++11

参考 C++11 规范 (5.1.2.13):

Refering to the C++11 specification (5.1.2.13):

出现在默认参数中的 lambda 表达式不应隐式或显式捕获任何实体.
[ 示例:

A lambda-expression appearing in a default argument shall not implicitly or explicitly capture any entity.
[ Example:

void f2() { int i = 1; void g1(int = ([i]{ return i; })()); // ill-formed void g2(int = ([i]{ return 0; })()); // ill-formed void g3(int = ([=]{ return i; })()); // ill-formed void g4(int = ([=]{ return 0; })()); // OK void g5(int = ([]{ return sizeof i; })()); // OK }

――结束示例 ]

但是，我们是否也可以使用 lambda 表达式本身作为函数参数的默认值?

However, can we also use a lambda-expression itself as the default value for a function argument?

例如

template<typename functor> void foo(functor const& f = [](int x){ return x; }) { }

推荐答案

是的.在这方面，lambda 表达式与其他表达式(例如 0)没有区别.但请注意，默认参数不使用推导.换句话说，如果你声明

Yes. In this respect lambda expressions are no different from other expressions (like, say, 0). But note that deduction is not used with defaulted parameters. In other words, if you declare

template<typename T> void foo(T = 0);

then foo(0); 将调用 foo 但 foo() 是格式错误的.您需要显式调用 foo() .由于在您的情况下您使用的是 lambda 表达式，因此没有人可以调用 foo ，因为表达式的类型(在默认参数的位置)是唯一的.但是你可以这样做:

then foo(0); will call foo<int> but foo() is ill-formed. You'd need to call foo<int>() explicitly. Since in your case you're using a lambda expression nobody can call foo since the type of the expression (at the site of the default parameter) is unique. However you can do:

// perhaps hide in a detail namespace or some such auto default_parameter = [](int x) { return x; }; template< typename Functor = decltype(default_parameter) > void foo(Functor f = default_parameter);

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