Question

law of non-contradiction 规定,两种相互矛盾的说法不能同时真实。这意味着表达方式

(a && !a)
(a == !a)
(a === !a)

始终应当评价一种错误价值,并且

(a || !a)

始终应当评估真相价值。

幸运的是,尽管如此,Javascript is a fun Language that allows You to do all ranges

What would be a good way to go about this?

Answer 1

我能做的最佳工作是:

[] == ![] // true

或

var a = []; 
a == !a

当然,这确实在做:[]=虚假/真实和! 这实际上只是技术性的。



EDIT: This is really a joke, but does w或k:

var a = false; var b = function() { return a = !a };
console.log(!!(b() && !b())); // true
console.log(b() == !b()); // true
console.log(b() === !b()); // true
console.log(b() || !b()); // true

Answer 2

这样做是为了:

var a =  0 ;
a == !a

(evaluates to true)

在此情况下,a = 虚假和! 伪造。

Answer 3

a=NaN;

var a=NaN, 
A=[(a && !a), (a == !a),(a === !a),(a || !a)];
alert(A)

/*  returned value: (Array)
NaN,false,false,true
*/

Answer 4

我仍然有没有发现任何突破<代码>&&和=,但此处为=和>:



Object.prototype.toString = function() {
    return false;
};
a = {};
b = (a || !a);
alert(a || !a);  //alerts false
alert(b == !b);  //alerts true

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