代理递归函数 [英] Proxying a recursive function

问题描述

想象一下一个简单的递归函数， jsdata-hide =falsedata-console =truedata-babel =false>

  //一个简单的递归函数.const count = n => n&& 1 + count（n-1）; //将代理中的函数包装到仪器输入和输出。函数仪器（fn）{return new Proxy（fn，{apply（target，thisArg，argumentsList）{console.log输入，... argumentsList）; const result = target（... argumentsList）; console.log（output，result）; return result;}}）;} //调用instrumented function.instrument（count） （2）;  

但是，这只记录输入和输出在最高层。我想找到一种方法，使 count 在调用时调用检测版本。

解决方案

函数调用 count ，这就是你需要包装的东西。您可以执行

  const count = instrument（n => n&& 1 + count（n-1） ）; 
  

或

  let count = n => n&& 1 + count（n-1）; 
 count = instrument（count）; 
  

对于其他一切，您需要将递归调用的函数动态注入到已检测函数中，类似于Y组合器的作用。

Imagine a simple recursive function, which we are trying to wrap in order to instrument input and output.

// A simple recursive function.
const count = n => n && 1 + count(n-1);

// Wrap a function in a proxy to instrument input and output.
function instrument(fn) {
  return new Proxy(fn, {
    apply(target, thisArg, argumentsList) {
      console.log("inputs", ...argumentsList);
      const result = target(...argumentsList);
      console.log("output", result);
      return result;
    }
  });
}

// Call the instrumented function.
instrument(count)(2);

However, this only logs the input and output at the topmost level. I want to find a way to have count invoke the instrumented version when it recurses.

解决方案

The function invokes count, so that is what you need to wrap. You can do either

const count = instrument(n => n && 1 + count(n-1));

or

let count = n => n && 1 + count(n-1);
count = instrument(count);

For everything else, you would need to dynamically inject the function for the recursive call into the instrumented function, similar to how the Y combinator does it.

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