"use strict"是否适用于在严格范围内声明的函数？

Question

我在jshint中得到这个,

[L16:C13] W034:不必要的指令"use strict".

不过我的问题是做这样的工作..

function () {
    "use strict";

    var a = function () {
        //stuff, (is this also strict)
    }

}

将东西严格吗？显然,严格声明之外的事情并不严格.但是其他函数内部是否开始非严格或继承严格性？

Answer 1

是的,这将是严格的; 而且,严格性是继承的.

任何任何子范围"use strict";也将是严格的.有没有必要,我可以看到声明的每个函数严格,如果你可以声明一个封装范围严格.

我尝试在范围内部和范围之外调用函数,以确保它没有任何区别.这是一个运行的测试mocha

var assert = require('assert');

var inside = function () {
    "use strict";
    var a = inside.a = function () {
        bar = 1;
        return bar;
    };
    var b = inside.b = function () {
        var bar = 1;
        return bar;
    };
    var c = inside.c = function () {
        "use strict";
        bar = 1;
        return bar;
    };
    var d = inside.d = function () {
        "use strict";
        var bar = 1;
        return bar;
    };

    describe('Inside Context', function () {
        it('inside a (no strict, no var)', a )
        it('inside b (no strict, var)', b )
        it('inside c (strict, no var)', c )
        it('inside d (strict, var)', d )
    } );

};

inside();

describe('Outside Context', function () {
    it('outside a (no strict, no var)', inside.a )
    it('outside b (no strict, var)', inside.b )
    it('outside c (strict, no var)', inside.c )
    it('outside d (strict, var)', inside.d )
} );

在这里我们看到内部和外部都没有用var声明变量我们有一个测试失败,表明strict正在工作:

Inside Context
  1) inside a (no strict, no var)  // fail
  ? inside b (no strict, var) 
  2) inside c (strict, no var)     // fail
  ? inside d (strict, var) 

Outside Context
  3) outside a (no strict, no var)  // fail
  ? outside b (no strict, var) 
  4) outside c (strict, no var)     // fail
  ? outside d (strict, var)