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

Question

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

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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

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

Answer 5

More elegant way: if "this" is object, convert it to true

"use strict"

var strict = ( function () { return !!!this } ) ()

if ( strict ) {
    console.log ( "strict mode enabled, strict is " + strict )
} else {
    console.log ( "strict mode not defined, strict is " + strict )
}

Answer 6

Another solution can take advantage of the fact that in strict mode, variables declared in eval are not exposed on the outer scope

function isStrict() {
    var x=true;
    eval("var x=false");
    return x;
}