javascript 有没有办法检查是否强制执行严格模式？
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Is there any way to check if strict mode is enforced?
提问by Deepak Patil
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
无论如何要检查是否强制执行严格模式“使用严格”，并且我们希望为严格模式执行不同的代码，为非严格模式执行其他代码。寻找类似的功能isStrictMode();//boolean
回答by ThiefMaster
The fact that thisinside a function called in the global context will not point to the global object can be used to detect strict mode:
this在全局上下文中调用的函数内部不会指向全局对象这一事实可用于检测严格模式：
var isStrict = (function() { return !this; })();
Demo:
演示：
> echo '"use strict"; var isStrict = (function() { return !this; })(); console.log(isStrict);' | node
true
> echo 'var isStrict = (function() { return !this; })(); console.log(isStrict);' | node
false
回答by Thalaivar
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
看起来你已经得到了答案。但是我已经写了一些代码。所以在这里
回答by noseratio
I prefer something that doesn't use exceptions and works in any context, not only global one:
我更喜欢不使用异常并且可以在任何上下文中工作的东西，而不仅仅是全局的：
var mode = (eval("var __temp = null"), (typeof __temp === "undefined")) ?
"strict":
"non-strict";
It uses the fact the in strict mode evaldoesn't introduce a new variable into the outer context.
它利用了严格模式下eval不会将新变量引入外部上下文的事实。
回答by Mehdi Golchin
Yep, thisis 'undefined'within a global method when you are in strict mode.
this是的'undefined'，当您处于严格模式时，它在全局方法中。
function isStrictMode() {
return (typeof this == 'undefined');
}
回答by Mehdi Golchin
More elegant way: if "this" is object, convert it to true
更优雅的方式：如果“this”是对象，则将其转换为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 )
}
回答by Yaron U.
Another solution can take advantage of the fact that in strict mode, variables declared in evalare not exposed on the outer scope
另一种解决方案可以利用在严格模式下声明的变量eval不会暴露在外部作用域中的事实
function isStrict() {
var x=true;
eval("var x=false");
return x;
}
