使用仿函数将 lambda 函数移植到 C03

C++中的 Lambda 函数是语法糖，为编写函子提供了非常简洁的语法。因此，通过将 lambda 函数转换为仿函数，可以在 C++ 03 中获得等效的功能（尽管更加冗长）：

// Some dummy types:
struct T1 {int dummy;};
struct T2 {int dummy;};
struct R {int dummy;};

// Code using a lambda function (requires C++11)
R use_lambda(T1 val, T2 ref) {
  // Use auto because the type of the lambda is unknown.
  auto lambda = [val, &ref](int arg1, int arg2) -> R {
    /* lambda-body */
    return R();
  };
  return lambda(12, 27);
}

// The functor class (valid C++03)
// Similar to what the compiler generates for the lambda function.
class Functor {
  // Capture list.
  T1 val;
  T2& ref;

public:
  // Constructor
  inline Functor(T1 val, T2& ref) : val(val), ref(ref) {}

  // Functor body
  R operator()(int arg1, int arg2) const {
    /* lambda-body */
    return R();
  }
};

// Equivalent to use_lambda, but uses a functor (valid C++03).
R use_functor(T1 val, T2 ref) {
  Functor functor(val, ref);
  return functor(12, 27);
}

// Make this a self-contained example.
int main() {
  T1 t1;
  T2 t2;
  use_functor(t1,t2);
  use_lambda(t1,t2);
  return 0;
}

如果 lambda 函数是 mutable，那么使 functor 的 call-operator 为非 const，即：

R operator()(int arg1, int arg2) /*non-const*/ {
  /* lambda-body */
  return R();
}