有谁知道 Javascript 中的 this 是什么意思？ `ǀǀ+(ǀ==ǀ-->ǀ+ǀ<--ǀ==ǀ)+ǀǀ`

Question

So I was looking at someones code on a website and while looking around, I saw this:

 var ǀ = 1; var ǀǀ = 21; var s = ǀǀ+(ǀ==ǀ-->ǀ+ǀ<--ǀ==ǀ)+ǀǀ; document.writeln(s);

This somehow equates to 42? Does anyone know how they are doing this? I thought || and | where for or ?

Answer 1

They're (ab)using a character that looks a whole lot like a vertical bar ( | ) such as what's used in "or" ( || ).

You could rewrite their code like this:

 var a = 1; var b = 21; var s = b + (a == a-- > a + a < --a == a) + b; document.writeln(s);

or further simplified:

 var s = 21 + (1 == 1 > 0 + 0 < -1 == -1) + 21; document.writeln(s); // Even further simplified var s = 21 + (true > false < true) + 21; document.writeln(s); // Clean up that middle case var s = 21 + (false) + 21; document.writeln(s); // Coerce to a number var s = 21 + 0 + 21; document.writeln(s); // Last simplification var s = 21 + 21; document.writeln(s);

No special operators. Just normal ones smushed together with odd-looking variable names.

Answer 2

ǀ and ǀǀ are (Unicode) variable names.

Space it out:

var ǀ = 1;
var ǀǀ = 21;
var s = ǀǀ + (ǀ == ǀ-- > ǀ + ǀ < --ǀ == ǀ) + ǀǀ;

Apply operator precedence and associativity:

var ǀ = 1;
var ǀǀ = 21;
var s = (ǀǀ + ((ǀ == ((ǀ-- > (ǀ + ǀ)) < --ǀ)) == ǀ)) + ǀǀ;

Apply order of evaluation:

var a = ǀǀ + ((ǀ == ((ǀ-- > (ǀ + ǀ)) < --ǀ)) == ǀ);
var s = a + ǀǀ;

var b = (ǀ == ((ǀ-- > (ǀ + ǀ)) < --ǀ)) == ǀ;
var a = ǀǀ + b;

var c = ǀ == ((ǀ-- > (ǀ + ǀ)) < --ǀ);
var b = c == ǀ;

var d = (ǀ-- > (ǀ + ǀ)) < --ǀ;
var c = ǀ == d;

var e = ǀ-- > (ǀ + ǀ);
var d = e < --ǀ;

var g = ǀ--;
var f = ǀ + ǀ;
var e = g > f;

Evaluate:

var ǀ = 1;
var ǀǀ = 21;

var g = ǀ--;
var f = ǀ + ǀ;
var e = g > f;
var d = e < --ǀ;
var c = ǀ == d;
var b = c == ǀ;
var a = ǀǀ + b;
var s = a + ǀǀ;

var ǀ = 0;
var ǀǀ = 21;

var g = 1;
var f = ǀ + ǀ;

var ǀ = 0;
var ǀǀ = 21;

var f = 0 + 0;
var e = g > f;

var ǀ = 0;
var ǀǀ = 21;

var e = 1 > 0;  // true
var d = e < --ǀ;

var ǀ = -1;
var ǀǀ = 21;

var d = true < -1;  // false
var c = ǀ == d;

var ǀ = -1;
var ǀǀ = 21;

var c = -1 == -1;  // true
var b = c == ǀ;

var ǀ = -1;
var ǀǀ = 21;

var b = true == -1;  // false
var a = ǀǀ + b;

var ǀ = -1;
var ǀǀ = 21;

var a = 21 + false;  // 21
var s = a + ǀǀ;

var ǀ = -1;
var ǀǀ = 21;

var s = 21 + 21;  // 42

Answer 3

It is a joke (a bad one). It isn't an operator. It is a collection of variable and operators put together to look intentionally confusing.

It is taking advantage of the visual similarities between ǀ (U+01C0 : LATIN LETTER DENTAL CLICK {pipe}) which is a character you can use in a variable name and | (U+007C : VERTICAL LINE {vertical bar}) which is an operator.

Answer 4

This is a mixture of obfuscation, and some abuse of weak typing.

First, ǀ and ǀǀ here are just variables with funny names - names that look a bit like, but aren't actually, the common character | , which would be an operator. So let's swap them out:

var a = 1;
var b = 21;
var s = b+(a==a-->a+a<--a==a)+b;
document.writeln(s);

Now we have various operations on one line, we need to work out the precedence of the different operators , so we can break it down in the right order. The operators we have, in order of highest to lowest precedence, are:

• ( ... ) to give a section higher precedence
• postfix --
• prefix --
• +
• > and < (left to right)
• == (multiple times, left to right)

So we can calculate the sum in this order:

• the part inside the brackets is the only interesting part, the rest is 21 + (something) + 21
• a-- gives 1 , but changes a to 0 ( b+(a==1>a+a<--a==a)+b )
• --a changes a to -1 and gives -1 ( b+(a==1>a+a<-1==a)+b )
• a is now -1 , so the very middle becomes -1+-1 ( b+(a==1>-1+-1<-1==a)+b ) which is -2 ( b+(a==1>-2<-1==a)+b )
• 1>-2 is true ( b+(a==true<-1==a)+b )
• true<-1 is false because true is treated as 1 ( b+(a==false==a)+b )
• a is -1 , but false is 0 , so a==false is like saying -1==0 , which is false ( b+(false==a)+b )
• false==a is again false ( b+(false)+b )
• treated as a number, false is 0, so b + 0 + b is 21 + 0 + 21 is 42

Interestingly, the in-place modification of a makes this open to various interpretations. For instance, it might be that all the unmodified a s are substituted as 1 (ie before the a++ ), giving some variations:

• with a as 1 , the very middle is 1+1 ( b+(a==1>1+1<-1==a)+b ) which is 0 ( b+(a==1>2<-1==a)+b )
• 1>2 is false ( b+(a==false<-1==a)+b )
• however, false<-1 is still false , so we get back to our original sequence