Name: _______________________

INEL 4206 Exam I

Fall 2009

September 29, 2009

Open book and notes. Only the text copies, instruction sheet handout, slide printouts and your own notes may be used. In the interests of originality and creativity please turn off all electronic communication devices including celulares, laptops, and pocket computing devices.

1. Describe and program a MIPS program equivalent to the C program

t1 = 1; s1=45; t2=10;
while{ t1 < s1)
t1 = t1 * t1;
t2 = t2 – 1;
s1 = s1 * t2 *(t1 – t2) + 1;
}

Assume t1, etc. will be in registers$ t1, etc. and s1 is in $s1 and do need to be initialized. You may need temporary registers.

Describe and program this loop in MIPS assembler (you do not need the .text and .data, etc., you only need to show the code fragment.

Please place your description here – this should include a description (it takes one short sentence) of what the program calculates

The program begins with the initialized definitions of t1, etc. The while loop, while t1<s1 starts with a label, again, followed by a test to see if t1<s1. If not, the loop is ended by a jump to a label done. The body of the loop is a straightforward translation of the three expressions to assembly; the longer expression takes a temporary to hold t1-t2 and several expressions

And your code here.

.text
li $t1,1
li $s1,45
li $t2,10
bge $t1,$s1,done
again: mul $t1,$t1,$t1
addui $t2,$t2,-1
sub $t4,$t1,$t2
mul $s1,$s1,$t2
mul $s1,$s1,$t4
addui $s1,$s1,1
b again
done:

2. Pseudoinstructions are expanded by the assembler into more than one machine instruction when instruction, as desired, is not part of the instruction set.

The symbolic assembler instruction
div $13, $12, S15

is expanded by SPIM into the sequence of absolute assembler instructions:
bne $15, $0, 8
break $0
div $12, $15
mflo $13

a. The bne instruction, if taken, skips the break instruction. Why is it done?

To catch the divide-by-zero case where the divisor, $15 is zero. Otherwise the break is bypassed and the division proceeds safely.

b. What does the mflo instruction do?

Places the quotient, which is in lo, into the destination $13

c. Suppose the instruction mflo is replaced by mfhi. What arithmetic operation is performed by this instruction sequence? Explain.

The remainder – is in hi after the division, is copied into $13 – this is the pseudoinstruction rem $13, $12, S15

3. These questions refer to various instruction set characteristics. Please answer briefly and avoid the dreaded RADQ.

a. Why is it necessary to use two instructions to load a 32-bit constant, as for example in the instruction li $t3,0x435312 into a register?

The RI format has room only for a 16-bit constant. So the lui loads the upper half, and the displacement (0x1152) is added to produce the 32-bit address.

b. Explain what the sequence of instructions
lui $at, 0x3101
lh $t3,0x1152($at) does

Loads the halfword starting at 0x31011152 into the word-size register $t3, sign extended.

c. Explain what the sequence of instructions (note these are the previous two in reverse order)

lh $t3,0x1152($at)
lui $at, 0x3101 does

Loads $t3 with the halfword at 0x????1152, sign-extended. The old value of $at becomes the upper half of the address – almost certainly a mistake.

d. Is the destination-first feature used for writing assembly code for SPIM a feature of the assembler in SPIM or is it true for all assemblers for the MIPS instruction set.? Explain.

No, some assemblers use the source-first convention. This is OK as long as you understand your assembler.

4. This question is based on the program file and the SPIM window shown on the next two pages. You may detach those two pages to make it easier to read them as you answer the questions.

a. Show C code for the loop – the lines starting at line 20 in the SPIM window (the break has to do with a breakpoint, but falls inside a pseudoinstruction)
You can use the notation $s1=s1 for variable names.

while (t0=v1$v0) {
if(t0==0) break;
v1=v0;
v0=t0;
}

b. When you reach a breakpoint and want to examine the code step-by-step do you answer yes or no to the continue question? What do you then do to step through the code?

You answer no, otherwise you would continue running until the next breakpoint and not necessarily see any of the steps. To step, you just keep stepping with the f10 key.

# Example of a leaf procedure

# Also illustrates system calls and little convenience functions

.data

.align 0

n1_prompt: .asciiz "First number - doesn't have to be the largest"

n2_prompt: .asciiz "Second number - doesn't have to be the smallest"

gcd_prompt: .asciiz "The gcd might just be "

newline: .asciiz "\n"

.text

.globl main

.align 2

get_int_prompted:

li $v0,4

syscall

li $v0,5

syscall

jr $ra

Euclid: move $v1,$a0

move $v0,$a1

again: rem $t0,$v1,$v0

beq $t0,$0,done

move $v1,$v0

move $v0,$t0

b again

done: jr $ra

main: la $a0,n1_prompt

jal get_int_prompted

beq $v0,$0,really_done

move $s0,$v0

la $a0,n2_prompt

jal get_int_prompted

move $a1,$v0

la $a0,gcd_prompt

li $v0,4

syscall

move $a0,$s0

jal Euclid

move $a0,$v0

li $v0,1

syscall

la $a0,newline

li $v0,4

syscall

b main

really_done:

li $v0,10

syscall # back to OS