为什么对定义函数的字符串的求值返回的是undefined而不是函数?

问题描述

function y(fct) {
   var a = 2;
   var fctStr = String(fct);
   var fct1 = eval(fctStr);
   console.log("fctStr=" + fctStr); // output: fctStr=function x() { return a + 1 }
   console.log("fct1=");
   console.log(fct1);  // output: undefined.  Why it is undefined?  I expect fct1 is a function.
   return fct1();  // exception: undefined is not a function.
}
function x() { return a + 1 }
y(x)   // I expect it returns 3.  However, it throws exception at above "return fct1()" statement.

Chrome中的此代码将获得fct1作为undefined.为什么?我希望fct1是一个函数.

This code in Chrome will get fct1 as undefined. Why? I expected that fct1 would be a function.

为什么要问这个问题是由于以下原因:

Why this question is asked is because of this: How to write function-y accepting parameter-fct_x which accesses var-a which is required to be defined in function-y?

推荐答案

您需要生成一个表达式.

You need an expression to be produced.

更改

var fct1 = eval(fctStr);

到

var fct1 = eval("("+fctStr+")");

这篇关于为什么对定义函数的字符串的求值返回的是undefined而不是函数?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！