不能有 auto 参数。您基本上有两个选项：

选项＃1：使用 std :: function p>

选项＃2：使用模板参数：

  template< typename F& ; 
 void f（F& lambda）{/ * ... * /} 
 

在某些情况下，选项＃2可能更有效，因为它可以避免嵌入lambda函数对象的潜在堆分配，但只有在 f 作为模板函数。它也可以增加编译时间和I-cache足迹，如同任何模板一样。注意，它可能没有效果，如果lambda函数对象足够小，它可以在 std :: function 对象中内联表示。

What is the right way to define a function that receives a int->int lambda parameter by reference?

void f(std::function< int(int) >& lambda);

or

void f(auto& lambda);

I'm not sure the last form is even legal syntax.

Are there other ways to define a lambda parameter?

You cannot have an auto parameter. You basically have two options:

Option #1: Use std::function as you have shown.

Option #2: Use a template parameter:

template<typename F>
void f(F &lambda) { /* ... */}

Option #2 may, in some cases, be more efficient, as it can avoid a potential heap allocation for the embedded lambda function object, but is only possible if f can be placed in a header as a template function. It may also increase compile times and I-cache footprint, as can any template. Note that it may have no effect as well, as if the lambda function object is small enough it may be represented inline in the std::function object.