如何使用“包装函数”调用不带参数n次的递归函数？

Question

So for a homework assignment we have to do the partial fraction definition of the golden ratio( n +(1/(n + 1/(.....) but we need to do it recursively. OK, not a problem I can do that.... but we need to call this recursive function 1000 times and cannot use parameter functions. The professor said to limit it to a finite number of calls we can use a wrapper function, but I have never heard of this term in class and do not know where to start on the wrapper function? Could someone help me out?

Please if you can tell me how to do it without sending me the solution to the partial fraction part itself, i'm doing poorly in this class and want to learn it on my own. I just need help with wrapper function. Thankyou!

Answer 1

A "wrapper function" is a function which wraps another function. It's essentially another function which calls the actual function.

As a simple and unrelated example, consider you want to get the maximum value of x and 10 . You could call std::max directly like value = std::max(x, 10) , but you don't want to pass the value 10 all the time, so you make a wrapper function:

int max10(int x)
{
     return std::max(x, 10);
}

The function max10 is a wrapper function.

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A possible solution for your problem is to not make the actual function recursive, but the wrapper function. The wrapper function will take one argument, n , which is the number of recursions it should make. If n is larger than zero then you call the actual function, and then make a recursive call to the wrapper function with the argument n - 1 .