在递归中意外的std :: bad_function_call

Question

我编写了以下代码,它使用类型的函数function<int(int)>.的功能compose,print,inc和guarded是其结合其他功能或产生一些外部作用助手.然后我用它们来构建我的程序:

/* start of the program */
function<int(int)> recursion();

function<int(int)> go =
  compose(guarded(10, recursion()), compose(inc, print("go")));

function<int(int)> recursion() {
  return compose(go, print("recursion"));
}

然而,打电话时recursion()(0),一个异常std::bad_function_call被抛出时go达到第二时间,但我不明白为什么.有悬挂参考还是空的std::function？此外,eta扩展go工作:

function<int(int)> go = [](int n) -> int {
  return compose(guarded(10, recursion()), compose(inc, print("go")))(n);
};

原始代码有什么问题？为什么替代方案有效？

完整代码:

#include <string>
#include <iostream>
#include <functional>

using namespace std;

/* helper functions, some combinators */

//composing two functions, f1 . f2
function<int(int)> compose(const function<int(int)>& f1, const function<int(int)>& f2) {
  return [f1,f2](int n) -> int {
    return f1(f2(n));
  };
}

function<int(int)> print(const string& msg) {
  return [msg](int n) -> int {
    cout << "print: " << msg << endl;
    return n;
  };
}

function<int(int)> inc = [](int n) -> int {
  cout << "in inc lambda: " << n << endl;
  return n+1;
};

//calls the given function `f` only when `n` is less then `m`
function<int(int)> guarded(int m, function<int(int)> f) {
  auto g = [m,f](int n) -> int { return n<m? f(n) : m; };
  return compose(g, print("guarded"));
}

/* start of the program */
function<int(int)> recursion();

function<int(int)> go =
  compose(guarded(10, recursion()), compose(inc, print("go")));

function<int(int)> recursion() {
  return compose(go, print("recursion"));
}

int main() {
  try {
    recursion()(0);
  } catch (bad_function_call e) {
    cout << "bad_function_call: " << e.what() << endl;
  }
  return 0;
}

Answer 1

在您的原始代码中,recursion()在初始化期间调用go.recursion本身试图使用的值go,但在那时go尚未初始化,导致问题.

在替代代码中,初始化go仅在go不调用的情况下分配lambda recursion.当recursion稍后被调用,go将已经被初始化.