Why does the evaluation of a string defining a function return undefined instead of the function?

问题

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.

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

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?

回答1:

You need an expression to be produced.

Change

var fct1 = eval(fctStr);

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

回答2:

Other's have answered in bits an pieces, but here's the whole answer:

function x() {return a + 1}

That defines a function named x, this is very similar to the following:

var x = function() {return a + 1}

In that case, an anonymous function is created and assigned to the variable x. So if you want to assign a function to a variable, there shouldn't be an identifier between function and ().

If you want the eval statement to return the function, then you need to wrap the function declaration with parenthesis ~~(and not give the function a name)~~. Edit: The function name is optional:

var x = eval('(function () {return a + 1})');

来源：https://stackoverflow.com/questions/27508784/why-does-the-evaluation-of-a-string-defining-a-function-return-undefined-instead

标签

javascript

eval