c++ lambda practice

First of all, you should read the lambda expression’s reference.

Lambda is a function

• different format, but the same.

    auto f1 = [](int i)->void {
        printf("%dn", i);
    };
  
    void f2(int i) {
        printf("%dn", i);
    }
  
    f1(1);
    f2(1);
  

  as a caller, f1 is the same with f2

• advantage: convenient

  • fastly define a simple function

  int main() {
    auto foo_min = [](int a, int b) -> int {
        return a < b ? a : b;
    };
    auto m = foo_min(1, 2);
  }
  

  • callback

  int do_read(..., int(*on_read)(const char* buf, int len))
  {
      ...
  }
  
  do_read(..., [](const char* buf, int len)->int{
      // handle buf ...
  });
  

Lambda is more than a function

• lambda is a closure. what is a closure?

  a closure is a record storing a function together with an environment:

  reference from wikipedia

• example for closure

    auto foo(int i) {
        return [i](int n)->void {
            printf("%dn", i + n);
        };
    }
  
    auto f1 = foo(1);
    auto f2 = foo(2);
  
    f1(1);
    f2(1);
  

• how about lambda meets c-style function type?

  • capture nothing

    typedef void (*func_t)();
    auto f = []()->void {
        printf("hellon");
    };
    

    f CAN be converted to func_t

  • capture something

    typedef void (*func_t)();
    int i = 5;
    auto f = [i]()->void {
        printf("%dn", i);
    };
    

    f CAN NOT be converted to func_t, because this lambda involves a closure that func_t CAN NOT express.

• what’s the type?

  speaking in code …

    typedef std::function<void(int)> the_type;
    std::vector<the_type> funcs;
  
    for(int i = 0; i < 5; ++i) {
        funcs.push_back([i](int n)->void{
            printf("%dn", i + n);
        });
    }
  
    for (auto f : funcs) {
        f(10);
    }
  

  output

    10
    11
    12
    13
    14