divpwr2 - Compute x/(2^n), for 0

Question

divpwr2 - Compute x/(2^n), for 0 <= n <= 30 * Round toward zero * Examples: divpwr2(15,1) = 7, divpwr2(-33,4) = -2 * Legal ops: ! ~ & ^ | + << >> * Max ops: 15 Can someone help me out with this please

e.mccormick · Answer

Well, in what language and what did you do so far?

anonymous · Answer

In C language ... i tried to use loops to finish it but my professor said i cannot use that in this as a coding rule ..

e.mccormick · Answer

I do not see why you would need a loop.

x/(2^n)  $\leftarrow$ That is the calculation you need
for 0 <= n <= 30 $\leftarrow$ That is the valid input for n, so you just need to make sure n does not fall outside that domain.
Round toward zero $\leftarrow$ simple enough.

Examples: divpwr2(15,1) = 7, divpwr2(-33,4) = -2 $\leftarrow$ OK, so lets look:

divpwr2(x,n)
divpwr2(15,1) = 15/(2^1) = 15/2 = 7.5 which rounds to 7
divpwr2(-33,4) = -33/(2^4) = -33/16 = -2.0625 which rounds... hmmm... They are saying it is -2 but tat does not seem right.  To me, "towards 0" would be round down above 0 and round up below 0.  But it looks like they are just using round off.