Why does the evaluation of a string defining a fun

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);

to

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})');