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def func_1(a,b):
   using_a = func_2(a-b, b) 
   with_b= func_3(lambda c: func_4(b, c), using_a)
   without_b = func_1(a, b-1)
   return func_5(with_b, without_b)

//except func_1 all other functions have already been initiated.

I'm not able to understand how c is computed in func4(b,c)

with_b= func_3(lambda c: func_4(b, c), using_a)

note without_b and the return uses recursion.

Here's the actual full code for context the syntax above refers to the partitions part of the code..

https://pastebin.com/iZ9HbzZ8

zram
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    It's still not clear what you are asking. Do you understand that `lambda c:` creates a new variable `c` which is local inside the `lambda`? Just like a function argument in a named function. – tripleee Sep 24 '22 at 06:51
  • Oh ok, so it creates the local variable c...and then passes it on to func4(b,c) as well? Right! Thanks.. – zram Sep 24 '22 at 06:53

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