Is there any way to check if strict mode is enforced?

Is there anyway to check if strict mode 'use strict' is enforced , and we want to execute different code for strict mode and other code for non-strict mode. Looking for function like isStrictMode();//boolean

The fact that this inside a function called in the global context will not point to the global object can be used to detect strict mode:

var isStrict = (function() { return !this; })();

Demo:

> echo '"use strict"; var isStrict = (function() { return !this; })(); console.log(isStrict);' | node
true
> echo 'var isStrict = (function() { return !this; })(); console.log(isStrict);' | node
false

I prefer something that doesn't use exceptions and works in any context, not only global one:

var mode = (eval("var __temp = null"), (typeof __temp === "undefined")) ? 
    "strict": 
    "non-strict";

It uses the fact the in strict mode eval doesn't introduce a new variable into the outer context.

function isStrictMode() {
    try{var o={p:1,p:2};}catch(E){return true;}
    return false;
}

Looks like you already got an answer. But I already wrote some code. So here

Yep, this is 'undefined' within a global method when you are in strict mode.

function isStrictMode() {
    return (typeof this == 'undefined');
}

Warning + universal solution

Many answers here declare a function to check for strict mode, but such a function will tell you nothing about the scope it was called from, only the scope in which it was declared!

function isStrict() { return !this; };

function test(){
  'use strict';
  console.log(isStrict()); // false
}

Same with cross-script-tag calls.

So whenever you need to check for strict mode, you need to write the entire check in that scope:

var isStrict = true;
eval("var isStrict = false");

Unlike the most upvoted answer, this check by Yaron works not only in the global scope.