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

please send me the code for multiplying sparse matrix using c

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