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I've got the following task:
Compute
x/(2^n), for0 <= n <= 30using bit shifting.Requirement: Round toward zero.
Examples:
divpwr2(15,1) = 7 divpwr2(-33,4) = -2Legal operators:
! ~ & ^ | + << >>Maximum number of operators: 15
Here is what I've got so far:
public int DivideByPowerOf2(int x, int n)
{
//TODO: find out why DivideByPowerOf2(-33,4) = -3 instead of -2
return x >> n;
}
DivideByPowerOf2(15,1) = 7 is ok.
But DivideByPowerOf2(-33,4) = -3 instead of -2. Why?
577
When shifting right with a logical right shift, the least-significant bit is lost and a 0 is inserted on the other end. For positive numbers, a single logical right shift divides a number by 2, throwing out any remainders.
To divide a number, a binary shift moves all the digits in the binary number along to the right and fills the gaps after the shift with 0: to divide by two, all digits shift one place to the right. to divide by four, all digits shift two places to the right. to divide by eight, all digits shift three places to the ...
Compute t = (N - D); . If (t >= 0) , then set the least significant bit of Q to 1, and set N = t . Left-shift N by 1. Left-shift Q by 1.
Pay close attention to the rounding behavior.
/ (integer divide) always rounds toward zero.
184
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