Answer / senthil

f returns (n * 2^(l-r))

n<<l => n * 2^l
(n<<l)>>r => (n * 2^l)/(2^r)=>(n * 2^(l-r))

Answer / guest

first it will call to n and then it will move on l or r, if
it is less then or equal to is there so it will return the
value ..... or else it will print the output

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