Re: [Haskell-beginners] Usage of $

s mp $ n - 1 parses as s mp (n - 1) s mp n - 1 parses as (s mp n) - 1 On Sun, Jun 7, 2015 at 3:09 PM, Shishir Srivastava < shishir.srivastava@gmail.com> wrote:

Hi,

Can someone please highlight how '$' is making the difference in the execution of this function. With $ in place the function runs fine but without $ for large input values the function overflows the stack.

-------- s mp 1 = 4 s mp n = ((s mp $ n-1)^2-2) `mod` mp --------

I cannot understand the difference between (s mp $ n-1) and (s mp n-1)

Thanks, Shishir Srivastava +44 (0) 750 127 5019

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