Assignment 10, due Apr 11Solutions
Part of
the homework for 22C:60 (CS:2630), Spring 2014

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_______________ _____________          s e  m 
There are no hidden bits here, and no special cases for e = 0000 or e = 1111.
a) What is the exact decimal representation of the largest positive number in this floating point system? (0.3 points)
01111111 = 0.111_{2} × 2^{111} = 7/8 × 2^{7} = 7 × 2^{4} = 112
b) What is the exact decimal representation of the most extreme negative number in this floating point system? (0.1 points)
11111111 = 112
c) What is the exact decimal representation of the smallest nonzero positive number in this floating point system? (0.3 points)
00000001 = 0.001_{2} × 2^{8} = 1/8 × 2^{8} = 0.00048828125
Note that we did not move the point when we added the hidden bit, so the hidden bit is in the one's place of the mantissa.
a) What is the exact decimal representation of the largest noninfinite positive number in this floating point system? (0.3 points)
01110111 = 1.111_{2} × 2^{110} = 1 7/8 × 2^{6} = 15 × 2^{3} = 120
b) What is the exact decimal representation of the most extreme noninfinite negative number in this floating point system? (0.1 points)
11110111 = 120
c) What is the exact decimal representation of the smallest nonzero positive number in this floating point system? (0.3 points)
00000001 = 0.001_{2} × 2^{7} = 2^{3} × 2^{7} = 2^{10} = 0.0009765625
d) What is the binary representation of 1.0 in this floating point system. (0.3 points)
01000000 = 1.000_{2} × 2^{0}
A problem: Write Hawk code to compute the distance, leaving the result in FPA0. Assume the floating point coprocessor is already turned on in singleprecision mode. (1.0 point)
In the following solution, the code has been optimized to give the coprocessor time to finish each operation. So, once it is busy computing on one accumulator, the next operation, if possible, involves the other acumulator. Toward the end, we have nothing else to do. A comment indicates where it would be good to do something else.
The assignment did not require or even suggest this optimization. There are many variations that are possible, involving reordering the earlier instructions here or using other registers. If you COGET each result immediately, you can use just one floating point accumulator, but you will need more general purpose registers.
LOADS R5,R3 COSET R5,FPA0 ;  fpa0 = p1.x LOADS R5,R4 COSET R5,FPSUB+FPA0 ;  fpa0 = fpa0  p1.x LOAD R5,R3,4 COSET R5,FPA1 ;  fpa1 = p1.y LOAD R5,R4,4 COSET R5,FPSUB+FPA1 ;  fpa1 = fpa1  p2.y COGET R5,FPA0 COSET R5,FPMUL+FPA0 ;  fpa0 = fpa0 ** 2 = (p1.x  p1.y)**2 COGET R5,FPA1 COSET R5,FPMUL+FPA1 ;  fpa1 = fpa1 ** 2 = (p1.y  p2.y)**2 COGET R5,FPA0 COSET R5,FPADD+FPA1 ;  fpa1 = fpa1 + fpa0 ;  can we do something here? COGET R5,FPA1 COSET R5,FPSQRT+FPA0 ; fpa0 = sqrt((p1.xp1.y)**2+(p1.yp2.y)**2)
The code in question is:
BITTST R3,KSTATRD BBR KBDPOLL ; } while ((kbdstat & kstatrd) == 0);Substituting the definition for KSTATRD this is:
BITTST R3,0 BBR KBDPOLL ; } while ((kbdstat & kstatrd) == 0);The BITTST macro in hawk.h translates this to:
ADDSR R0,R3,1 BCR KBDPOLLThis can also be described as:
SR R3,1 BCR KBDPOLL