How can I detect I'm inside an eval() call?

Question

Does there exist a string s such that

(new Function(s))();

and

eval(s);

behave differently? I\'

Answer 1

if (new Error().stack.indexOf('at eval') > -1) {
    console.log('Oh noo, I am being evaled');
}

Answer 2

Check for the arguments object. If it exists, you're in the function. If it doesn't it has been evaled.

Note that you'll have to put the check for arguments in a try...catch block like this:

var s = 'try {document.writeln(arguments ? "Function" : "Eval") } catch(e) { document.writeln("Eval!") }';
(new Function(s))();
eval(s);

Demo

Solution to nnnnnn's concern. For this, I've edited the eval function itself:

var _eval = eval;
eval = function (){
    // Your custom code here, for when it's eval
    _eval.apply(this, arguments);
};

function test(x){
    eval("try{ alert(arguments[0]) } catch(e){ alert('Eval detected!'); }");
}
test("In eval, but it wasn't detected");​

Answer 3

The current answer does not work in strict mode since you can't redefine eval. Moreover, redefining eval is problematic for many other reasons.

The way to differenciate them is based on the fact that well... one of them creates a function and what doesn't. What can functions do? They can return stuff :)

We can simply exploit that and do something with return:

// is in function
try {
     return true;
} catch(e) { // in JS you can catch syntax errors
     false; //eval returns the return of the expression.
}

So in example:

var s = "try{ return true; }catch(e){ false; }";
eval(s); // false
Function(s)(); // true
(new Function(s))(); // true, same as line above
(function(){ return eval(s); })(); // the nested 'problematic' case - false