问题描述
如何递归地实现lambas。
how to implement lambas recursively.
推荐答案
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.
Func<int, int> factorialValue = null;
factorialValue = x => (x == 0) ? 1 : x * factorialValue(x - 1);
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