c++ - 如何在 C++ 函数中实现递归函数？

Question

我了解 lambda 函数的工作原理。问题在于，程序在编译器推断出“auto”应该是什么之前调用了函数 recursiveFunction()。问题是，它是一个递归函数，因此函数本身位于定义中。

#include <iostream>
using namespace std;

template <class T>
class Class {
    public:
        int foo(int x);
};

template <class T>
int Class<T>::foo(int x) {
    auto recursiveFunction = [=](int n)->int {
        if (n <= 1) return 1;
        else return n*recursiveFunction(n-1);
    };
    return recursiveFunction(x);
}

int main() {
    Class<int> c;
    cout << c.foo(5) << endl;
    return 0;
}

我还使用使用模板的类来实现这一点，以防影响问题。

错误消息如下:

main.cpp: In instantiation of 'int Class<T>::foo(int) [with T = int]':
main.cpp:21:20:   required from here
main.cpp:14:40: error: use of 'recursiveFunction' before deduction of 'auto'
         else return n*recursiveFunction(n-1);

谢谢!

Answer 1

已回复 here :

The second snippet runs into [dcl.spec.auto]/10:

If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed.

需要 foo 的类型来确定 lambda 体内表达式 foo 的类型，但此时您还没有推导出 foo 的类型，因此程序格式错误。

更多引用:

• lambda capture during initialization should be an error
• "a variable declared with an auto specifier cannot appear in its own initializer"

修复:https://godbolt.org/z/np3ULe

#include <iostream>
#include <functional>

template <class T>
class Class {
 public:
  int foo(int x);
};

template <class T>
int Class<T>::foo(int x) {
  std::function<int(int)> fac = [&fac](int n) -> int {
    if (n <= 1)
      return 1;
    else
      return n * fac(n - 1);
  };
  return fac(x);
}

int main() {
  Class<int> c;
  std::cout << c.foo(5) << std::endl;
  return 0;
}