Question

For each of the following, describe how to implement the operation of the variable "int x" using only bitwise, addition, and subtraction operators (<<, >>, +, -, &, |, ^, ~)

A) x * 2

B) x * 14

C) x * -44

D) x * 1023

Problem 2. Floating point encoding. Write out the 32 bit IEEE floating point binary representation for each of the following numbers:

A) 1.0

B) -2.0

C) -1000000.00

D) nan

E) -inf

F) 3.402823e38

Answer 1

A)
x<<1

B)
x<<4 - x<<1

C)
x<<4 - x<<6 + x<<2

D)
x<<10 - x

A)
1.0 in simple binary => 1.0
so, 1.0 in normal binary is 1.0 => 1. * 2^0

single precision:
--------------------
sign bit is 0(+ve)
exp bits are (127+0=127) => 01111111
frac bits are 00000000000000000000000

so, 1.0 in single-precision format is 0 01111111 00000000000000000000000
Answer: 0 01111111 00000000000000000000000

B)
-2.0 in simple binary => 10.0
so, -2.0 in normal binary is 10.0 => 1. * 2^1

single precision:
--------------------
sign bit is 1(-ve)
exp bits are (127+1=128) => 10000000
frac bits are 00000000000000000000000

so, -2.0 in single-precision format is 1 10000000 00000000000000000000000
Answer: 1 10000000 00000000000000000000000

C)
-1000000.0 in simple binary => 11110100001001000000.0
so, -1000000.0 in normal binary is 11110100001001000000.0 => 1.1110100001001 * 2^19

single precision:
--------------------
sign bit is 1(-ve)
exp bits are (127+19=146) => 10010010
frac bits are 11101000010010000000000

so, -1000000.0 in single-precision format is 1 10010010 11101000010010000000000
Answer: 1 10010010 11101000010010000000000

D)
nan
Answer: 01111111111111111111111111111111

E)
-inf
Answer: 11111111100000000000000000000000

F)
3.402823e38
3.402823e+38 in simple binary => 11111111111111111111110010110011010101101100100100100000000000000000000000000000000000000000000000000000000000000000000000000000
so, 3.402823e+38 in normal binary is 11111111111111111111110010110011010101101100100100100000000000000000000000000000000000000000000000000000000000000000000000000000 => 1.111111111111111111111 * 2^127

single precision:
--------------------
sign bit is 0(+ve)
exp bits are (127+127=254) => 11111110
frac bits are 11111111111111111111100

so, 3.402823e+38 in single-precision format is 0 11111110 11111111111111111111100
Answer: 0 11111110 11111111111111111111100