“如何使用严格”修改“this”的规则在Javascript？ [英] How does "use strict" modify the rules for "this" in Javascript?

问题描述

我试图理解使用严格的这个的规则;修改如下。

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

阅读后（ http://unschooled.org/2012/03/understanding-javascript-this/ ）我最好的猜测是，因为functon isStrictModeOn（）没有附加到任何东西，所以这是指null 。这被认为是一个更合理的替代Javascript只是将它附加到全局对象。在这种情况下，使用严格的变更是否正确解释？

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.

推荐答案

这几乎是正确的。在严格模式下，当一个函数在没有接收器的情况下被调用时，这个是 undefined （不是空）。该函数的更好版本是：

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';
}

这类函数的一个固有问题是严格性是词汇决定的，像范围一样，所以它是静态的。包含其自己的use strict; 的测试器功能不是很有用;它实际上只告诉您JavaScript运行时是否理解严格模式。一个没有自己的use strict; 告诉你它所定义的词汇上下文是否处于严格模式。那就是：

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();
}

会在通话时告诉您是否 use strict; 对于定义函数的范围有效。我想这可能有用。但是，如果对该函数的引用泄漏到其严格性不同的其他上下文中，它将继续在其定义时报告其静态严格性。

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.

就个人而言，我会选择通过在最外层调用use strict; 来确保我的代码绝对处于严格模式。这样就没有必要检查它了。

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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