javascript - Is 'use strict' implied when ES modules is used? - Stack Overflow

admin管理员组

文章数量:1406063

I was learning JS and came across 'use strict'. Then, here Should I 'use strict' for every single javascript function I write? @Bergi says that "you only should put 'use strict' in the module scope - once per file - so that it is inherited by all your functions". Is it true that by "it is inherited by all your functions" means each function in a certain module uses 'use strict' inside of itself?

I was learning JS and came across 'use strict'. Then, here Should I 'use strict' for every single javascript function I write? @Bergi says that "you only should put 'use strict' in the module scope - once per file - so that it is inherited by all your functions". Is it true that by "it is inherited by all your functions" means each function in a certain module uses 'use strict' inside of itself?

• javascript

Share Improve this question Follow asked Mar 5, 2020 at 10:08 WertuWertu 13766 bronze badges 4

• ecma-international/ecma-262/6.0/#sec-strict-mode-code – George Commented Mar 5, 2020 at 10:12
• 2 any type="module" script is automatically in strict-mode, yes. – ASDFGerte Commented Mar 5, 2020 at 10:13
• @ASDFGerte, so e.g if we have function like function Addition(){'use strict'}, each function inside es module has use strict in itself. Right? – Wertu Commented Mar 5, 2020 at 10:22
• 1 If you are using native JavaScript modules, then all the code within your modules will be in strict mode by default. If you are using older hand-rolled modularisation patterns, then you will need to declare 'use strict' at the top of the module. – Ben Aston Commented Mar 5, 2020 at 10:23

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 8

Is it true that by "it is inherited by all your functions" means each function in a certain module uses 'use strict' inside of itself?

A function will run in strict mode if:

(1) The function is inside an ES6 module (these all run in strict mode regardless), or

(2) There is a proper 'use strict' directive above the function definition, lexically.

In addition to a few less mon instances, such as inside a constructor.

The way 'use strict' is inherited lexically works the same way variables can be referenced, and how variable scope works - if any outer block of a function is strict, then the inner function is strict as well. So, for example, with the following code:

'use strict';
function foo() {
  function bar() {
    function baz() {
    }
  }
}

No matter what other code you put inside foo, bar, and baz, all of those functions, no matter how they're run, will run in strict mode. If one of these function is called from a source which isn't running in strict mode, the called function will still run in strict mode. For example:

<script>
'use strict';
function foo() {
  return function bar() {
    return function baz() {
      console.log('baz running. strict:', this === undefined);
    }
  }
}
foo()()();
</script>
<script>
// Non-strict script calling strict function:
foo()()();
</script>

本文标签： javascriptIs 39use strict39 implied when ES modules is usedStack Overflow