A Solution to Homework Set 6

Since homework 6 will not be collected and graded, it might be helpful to have a solution to study. Please first try to solve the problem on your own.

Problem 1

We have a binary value: 11111110111011011111101011001110

To express this in octal, first break it into groups of 3 bits, starting the grouping from the Least Significant Bit (the rightmost bit):

11 111 110 111 011 011 111 101 011 001 110
 3   7   6   7   3   3   7   5   3   1   6

To express this in hexadecimal, break it into groups of 4 bits, starting the grouping from the Least Significant Bit:

1111 1110 1110 1101 1111 1010 1100 1110
   F    E    E    D    F    A    C    E

Problem 2

The least significant bit of the "direction" is at bit 6 of gps. In the result, this must be bit 0. So we can begin by shifting to the right 6 places:

(gps >> 6)

However, only bits 0—5 of this are the direction; there may be bits to the left of bit 5 that are other "fields" in gps. So we need to create a mask that will only keep the right-most 6 bits:

0011 1111
   3    F

So our final result is:

dir = (gps >> 6) & 0x3F;