Obfuscation example

obfuscated.clj

; first I received this, and its result is 5. why?
((fn [[?<& %*+ $*|]](?<& %*+ $*|))({['*$] [+ (+ (*) (*)) (* 3 (*))]} ['*$]))

; so I formatted the code
((fn [[?<& %*+ $*|]] 
  (?<& %*+ $*|))

({['*$] [+ (+ (*) (*)) (* 3 (*))]} ['*$]))

; ?<& is the first parameter, %*+ is the second and $*| is the last. let's rename to x, y and z
((fn [[x y z]] 
  (x y z))

({['*$] [+ (+ (*) (*)) (* 3 (*))]} ['*$]))

; '*$ is an id of a hash item. let's call it 'id

((fn [[x y z]] 
  (x y z))

({['id] [+ (+ (*) (*)) (* 3 (*))]} ['id]))

; if you evaluate (+ (*) (*)) alone, you will get 2, because (*) is 1, so (+ 1 1) equals 2
((fn [[x y z]] 
  (x y z))

({['id] [+ 2 (* 3 (*))]} ['id]))

; already knowing that (*) is 1, (* 3 (*)) means (* 3 1) that is equals 3
((fn [[x y z]] 
  (x y z))
({['id] [+ 2 3]} ['id]))

; let's attack the hash. we know that ({:id :value} :id) is :value, so ({['id] [+ 2 3]} ['id]) is [+ 2 3]
((fn [[x y z]] 
  (x y z)) [+ 2 3])

; in the end, using destructuring, you're passing + as parameter to z, 2 to y and 3 to z, so you have (+ 2 3), that is 5
(+ 2 3)

5