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
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7 Answers
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
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3For clarification, the return statement is equivalent to `return this === undefined`, it's not comparing it to the global object, it's just checking if `this` exists. – aljgom Mar 15 '17 at 21:40
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.
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Just out of curiosity, how bulletproof is this in 2015, now that ES6 is here? – John Weisz Jul 18 '15 at 17:02
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3I verify that it works in ES6 on latest chrome and nodejs. – Michael Matthew Toomim Nov 30 '16 at 00:27
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1
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
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1This is better than Mehdi's answer as it will work everywhere, not only in a global scope. Upped. :) – mgol Aug 15 '12 at 23:45
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7This results in a syntax error, which happens before the code runs, so it can't be caught... – skerit Jun 28 '13 at 12:11
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5This will not work in ES6 either as the check is removed to allow computed property names. – billc.cn Jan 30 '15 at 10:22
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@skerit Can you elaborate on your syntax error? I do not get one. – Robert Siemer Feb 25 '20 at 00:48
Yep, this
is 'undefined'
within a global method when you are in strict mode.
function isStrictMode() {
return (typeof this == 'undefined');
}
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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 )
}
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.
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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;
}
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