CZ1101, Tutorial 6, Answers

The question is that given a simple computer (we built in Chapter 5) which has only nine instructions lw, sw, add, sub, and, or, slt, beq, and j, can we do many of the other operations (in software) not found in the instruction set?
(a) We can do (bne $4, $5, L) by
```
   beq $4, $5, Tmp
   j L               # if $4 != $5, jump to L
Tmp:                 # if $4 == $5, jump here, nothing happens
```
(b) We can do a shift left 1 bit (sll $2, $3, 1) by,
```
   add $2, $3, $3    # shift left 1 bit is the same as multiplying by 2
```
Shift right 1 bit (srl $2, $3, 1)? It's doable. But I have not seen a simple way. But we must assume that we can initialize the memory in arbitrary way. Then we can mask out the i-th bit, if it is a one we set the (i-1)-th bit as one, using logical AND and OR instructions.
(c) Can we program it to do multiplication and division with this computer? The answer is yes.
(d) Floating-point arithmetic? Yes.

The binary numbers represent the following instruction sequence:

 (1) 0x2002ffff  addi $2, $0, -1              
 (2) 0x3043000f  andi $3, $2, 15             
 (3) 0x14400003  bne $2, $0, 3
 (4) 0x00031842  srl $3, $3, 1                
 (5) 0x0810000f  j 0x0010000f
 (6) 0xafa30000  sw $3, 0($29)              
 (7) 0x00621022  sub $2, $3, $2            
 (8) 0x3c043f80  lui $4, 16256            
 (9) 0xafa40004  sw $4, 4($29)           
(10) 0xc7a00004  lwc1 $f0, 4($29)       
(11) 0x3c044080  lui $4, 16512         
(12) 0xafa40008  sw $4, 8($29)        
(13) 0xc7a20008  lwc1 $f2, 8($29)    
(14) 0x46020103  div.s $f4, $f0, $f2

(b) The values changed in each of the instructions are (1) $2 = -1, (2) $3 = 15, (3) jump forward by a distance of 3 (skipping srl and j, spim and MIPS differ on this), (6) save 15 on 0($29), (7) $2 = 16, (8) $4 = 0x3f800000, (9) save 0x3f800000 in 4($29), (10) $f0 = 0x3f80000 = 1.0, (11) $4 = 0x40800000, (12) save 0x40800000 in 8($29), (13) $f2 = 0x40800000 = 4.0, (14) $f4 = 1/4 = 0.25 = 0x3e800000.