The rules for completing lab include:

All of these problems can be done with the eight operators:

! ~ & ^ | + << >>

and some require you to stick with just a subset of these.

You are also limited to constants with length at most 8-bits

i.e. you are limited to constants with two hexadecimal digits.

You must use straight-line code no loops or conditionals.

Each puzzle also has a rating of 1 (easiest) to 4 (hardest).

There is a limit on the number of operations you may use (just to eliminate brute force solutions).

Each puzzle is worth 5 points

Assignment:

Puzzle Six byteSwap - swaps the nth byte and the mth byte

Examples: byteSwap(0x12345678, 1, 3) = 0x56341278

byteSwap(0xDEADBEEF, 0, 2) = 0xDEEFBEAD

You may assume that 0 <= n <= 3, 0 <= m <= 3

Legal ops: ! ~ & ^ | + << >>

Max ops: 40

Rating: 2

int byteSwap(int x, int n, int m) { }

Puzzle Seven isPositive - Return 1 if x>0, return 0 otherwise

Difficulty Rating - 3

Legal Operators: ! ~ & ^ | + << >>

Max Ops: 12

isPositive(int x) { }

Puzzle Eight isLess - if x < y then return 1, else return 0

Example: isLess(4,5) = 1.

Legal ops: ! ~ & ^ | + << >>

Max ops: 30

Rating: 3

isLess(int x, int y) { }

Puzzle Nine threeFourths - multiplies by 3/4 rounding toward 0,

Should exactly duplicate effect of C expression (x*3/4), including overflow behavior.

Examples: ezThreeFourths(11) = 8

ezThreeFourths(-9) = -6

ezThreeFourths(1073741824) = -268435456 (overflow)

Legal ops: ! ~ & ^ | + << >>

Max ops: 12

Rating: 3

int threeFourths(int x) { }

Puzzle Ten bang - Compute !x without using !

Examples: bang(3) = 0, bang(0) = 1

Legal ops: ~ & ^ | + << >>

Max ops: 20

Rating: 4

int bang(int x) { }

Puzzle Eleven (Extra Credit) bitCount - returns count of number of 1's in word

Examples: bitCount(5) = 2, bitCount(7) = 3

Legal ops: ! ~ & ^ | + << >>

Max ops: 40

Rating: 4