How Does [b][b = A,0] Swap Between A And B?
As gdoron pointed out, var a = 'a'; var b = 'b'; a = [b][b = a,0]; Will swap a and b, and although it looks a bit of hacky, it has triggered my curiosity and I am very curious a
Solution 1:
var a = "a";
var b = "b";
a = [b][b = a, 0];
Let's break the last line into pieces:
[b] // Puts b in an array - a safe place for the swap.
[b = a] // Assign a in b
[b = a,0] // Assign a in b and return the later expression - 0 with the comma operator.
so finally it is a =[b][0]
- the first object in the [b]
array => b
assigned to a
read @am not I am comments in this question: When is the comma operator useful? It's his code...
Solution 2:
It might help (or hinder) to think of it terms of the semantically equivalent lambda construction (here, parameter c
takes the place of element 0):
a =(function(c){ b = a; returnc; })(b);
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