递归地实现lambdas [英] implement lambdas recursively

问题描述

如何递归地实现lambas。

how to implement lambas recursively.

推荐答案

解决方案1非常好但是它留下了一个漏洞的错误。

这取决于递归调用绑定到名称 factorialValue 。

如果被泄露它就会崩溃。

非常人为的例子：

Solution 1 is pretty good but it leaves a vulnerability for error.
It depends on the binding to the name factorialValue for the recursive call.
If were compromised it falls apart.
Very contrived example:

Func<int, int> Factorial = null;
Factorial = x => (x <= 0) ? 1 : x * Factorial(x - 1);
Func<int, int> incorrectFactorial = Factorial;
Factorial = x => x;

int ans = incorrectFactorial(5);

由于重新分配了标识符<$ c $，因此 ans 的值为20 c> Factorial 。

绑定递归方法的名称需要保护：

The value of ans here is 20 because of the reassignment of the identifier Factorial.
The name binding the recursive method needs to be protected:

Func<int, int> Factorial = null;
Factorial = x => {
  Func<int, int> internalFactorial = null;
  internalFactorial = xi => (xi <= 0) ? 1 : xi * internalFactorial(xi - 1);
  return internalFactorial(x);
};

Func<int, int> otherFactorial = Factorial;
Factorial = x => x;

int ans = otherFactorial(5);

这里 ans 的值是正确的120，因为递归标识符是范围内部的of lambda。

另一种方法是使用一个方法返回 Func< int，int> ：

Here the value of ans is the correct 120 because the recursion identifier was internal to the scope of the lambda.
Another way to accomplish this would be to have a method that returns a Func<int, int>:

static Func<int, int> GetFactorialFunction()
{
  Func<int, int> Factorial = null;
  Factorial = x => (x <= 0) ? 1 : x * Factorial(x - 1);
  return Factorial;
}

并调用返回的函数：

and invoke the returned function:

int ans = GetFactorialFunction()(5);

Of当然，此时使用lambda 可能是多余的，因为函数可以直接实现为命名的递归函数。

Of course, at this point using a lambda is probably superfluous since the function could be implemented directly as a named, recursive function.

可以递归地实现lambda。

试试这个例子

Its possible to implement lambda recursively.
Try this as an example

Func<int, int> factorialValue = null;
factorialValue = x => (x == 0) ? 1 : x * factorialValue(x - 1);

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