Pregunta

I'm trying to understand what rule for "this" that "use strict"; modifies in the below case.

After reading (http://unschooled.org/2012/03/understanding-javascript-this/) my best guess is that since the functon isStrictModeOn() is not "attached" to anything, this refers to null. Which is suppose to be a more sensible alternative to Javascript just attaching the this to the global object. Is that the correct interpretation of the change that "use strict" is making in this case?

http://www.novogeek.com/post/ECMAScript-5-Strict-mode-support-in-browsers-What-does-this-mean.aspx

function isStrictMode(){
    return !this;
} 
//returns false, since 'this' refers to global object and '!this' becomes false

function isStrictModeOn(){   
    "use strict";
    return !this;
} 
//returns true, since in strict mode, the keyword 'this' does not refer to global object, unlike traditional JS. So here,'this' is null and '!this' becomes true.
¿Fue útil?

Solución

That's almost correct. In strict mode, when a function is invoked without a receiver then this is undefined (not null). A better version of that function would be:

function isStrict() {
  "use strict";
  return (typeof this) === 'undefined';
}

An inherent problem with functions like that is that "strictness" is determined lexically, like scope, so it's static. A tester function that includes its own "use strict"; isn't very useful; it really only tells you whether the JavaScript runtime understands strict mode. One without its own "use strict"; tells you whether the lexical context in which it's defined is in strict mode. That is:

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

will tell you, when called, whether a "use strict"; was in effect for the scope at which the function is defined. I guess that could be useful. However, if a reference to that function "leaks" into some other context whose "strictness" differs, it's going to continue to report on its static strictness at the point of its definition.

Personally, I would opt for simply ensuring that my code is definitely in strict mode by invoking "use strict"; at the outermost layer possible. That way there's really no need to check for it.

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