Islamic University – Gaza
Engineering Faculty
Department of Computer Engineering
ECOM 3010: Computer Architecture Discussion
Chapter 2
Exercises with solutions
Eng. Eman R. Habib
October, 2013
2 Computer Architecture Discussion
Discussion exercises
Exercise 1:
Convert the following C statements to equivalent MIPS assembly language. Assume that the variables
f, g, I and j are assigened to registers $s0, $s1, $s2 and $s3 respectively. Assume that the base
address of the array A and B are in registers $s6 and $s7 respectively.
a) f = g + h + B[4]
lw $t0, 16($s7)
add $s0, $s1, $s2
add $s0, $s0, $t0
b) f = g – A[B[4]]
lw $t0,16($s7)
sll $t1, $t0, 2
add $t2, $t1, $s6
lw $t3, 0($t2)
sub $s0, $s1, $t3
Exercise 2: (2.4 from book)
Why doesn’t MIPS have a subtract immediate instruction?
Since MIPS includes add immediate and since immediate can be positive or negative, its range
2
15
, add immediate with a negative number is equivalent to subtract immediate with positive
number, so subtract immediate would be redundant.
Exercise 3: (2.6 from book)
Find the shortest sequence of MIPS instructions that extracts a field for the constant values i=5
and j=22 from register $t3 and places it in register $t0.
sll $t0, $t3, 9
srl $t0, $t0, 15
3 Computer Architecture Discussion
31-j = 31-22 = 9
sll $t0, $t3, 9
i+1+9 = 5+1+9 = 15
srl $t0, $t0, 15
Another solution:
i+1 = 6
srl $t0, $t3, 6
j-i = 22-5=17
andi $t0, $t0, 0x0001FFFF
Exercise 4: (2.32 from book)
Show the single MIPS instruction for this C statement:
b = 25 | a;
ori $t1, $t0, 25
Exercise 5:
Convert the MIPS instruction to machine language:
srl $s1, $t2, 3
srl is R-type, opCode is 0 and function is 2
$s1 = 17 is rd
$t2 = 10 is rt
rs unused
shamt is 3
0000 00/00 000/0 1010 /1000 1/000 11/00 0010 = 0x00A88C2
Op(6) rs(5) rt(5) rd(5) shamt(5) func(6)
4 Computer Architecture Discussion
Exercise 6:
Translate the following machine code to MIPS:
1010 11/10 000/0 1011 /0000 0000 0000 0100
43 16 11 4
Op(6) rs(5) rt(5) immediate(16)
Op = 43 = sw
16 = $s0
11 = $t3
sw $t3, 4($s0)
Exercise 7: (2.37 from book)
For each pseudoinstruction in the following table, produce a minimal sequence of actual MIPS
instructions to accomplish the same thing. In the following table, big refers to a specific
number that requires 32 bits to represent and small to a number that can fit in 16 bits.
Pseudoinstruction What it accomplishes Solution
move $t1, $t2 $t1 = $t2 add $t1, $t2, $zero
clear $t0 $t0 = 0 add $t0, $zero, $zero
beq $t1, small, L if($t1 == small)go to L addi $t0, $zero, small
beq $t1, $t0, L
beq $t2, big, L if($t2 == big)go to L lui $t1, upper(big)
ori $t1, $t1, lower(big)
beq $t2, $$t1, L
li $t1, small $t1 = small addi $t1, $zero, small
li $t2, big $t2 = big lui $t2, upper(big)
ori $t2, $t2, lower(big)
ble $t3, $t5, L if ($t3 <= $t5) go to L slt $t0, $t5, $t3 beq $t0, $zero, L bgt $t4, $t5, L if ($t4 > $t5) go to L slt $t0, $t5, $t4
bne $t0, $zero, L
bge $t5, $t3, L if ($t5 >= $t3) go to L slt $t0, $t5, $t3
beq $t0, $zero, L
addi $t0, $t2, big $t0 = $t2 + big lui $t1, upper(big)
ori $t1, $t1, lower(big)
add $t0, $t2, $t1
lw $t5, big($t2) $t5 = Memory[$t2 + big] lui $t1, upper(big)
ori $t1, $t1, lower(big)
add $t1, $t1, $t2
lw $t5, 0($t1)
5 Computer Architecture Discussion
Exercise 8:
Convert the following C fragment to equivalent MIPS assembly language. Assume that the
variables a and b are assigened to registers $s0 and $s1 respectively. Assume that the base
address of the array D is in register $s2.
while(a < 10){ D[a] = b + a; a += 1; } Loop: stli $t0, $s0, 10 beq $t0, $zero, exit sll $t1, $s0, 2 add $t1, $t1, $s2 add $t2, $s1, $s0 sw $t2, 0($t1) addi $s0, $s0, 1 j Loop exit: Exercise 9: Show the effects on memory and registers of the following instructions. Suppose a portion of memory contains the following data Address Data 0x10000000 0x12345678 0x10000004 0x9ABCDEF0 And register $t0 contains 0x10000000 and $s0 contains 0x01234567. Assume each of the following instructions is executed independently of the others, starting with the values given above. Hint: Don’t forget that the MIPS architecture is Big-Endian. The memory: Address Data 0x10000000 0x12 0x10000001 0x34 0x10000002 0x56 0x10000003 0x78 0x10000004 0x9A 0x10000005 0xBC 0x10000006 0xDE 0x10000007 0xF0 6 Computer Architecture Discussion a) lw $t1, 0($t0) $t1 = 0x12345678 b) lw $t2, 4($t0) $t2 = 0x9ABCDEF0 c) lb $t3, 0($t0) $t3 = 0x00000012 d) lb $t4, 4($t0) $t4 = 0xFFFFFF9A lb is sign extended e) lb $t5, 3($t0) $t5 = 0x00000078 f) lh $t6, 4($t0) $t6 = 0XFFFF9ABC lh is sign extended g) sw $s0, 0($t0) at address 0x10000000 will contain 0x01234567 Address Data 0x10000000 0x01 0x10000001 0x23 0x10000002 0x45 0x10000003 0x67 0x10000004 0x9A 0x10000005 0xBC 0x10000006 0xDE 0x10000007 0xF0 h) sb $s0, 4($t0) the address 0x10000004 will contain 0x67BCDEF0 Address Data 0x10000000 0x12 0x10000001 0x34 0x10000002 0x56 0x10000003 0x78 0x10000004 0x67 0x10000005 0xBC 0x10000006 0xDE 0x10000007 0xF0 7 Computer Architecture Discussion i) sb $s0, 7($t0) the address 0x10000004 will contain 0x9ABCDE67 Address Data 0x10000000 0x12 0x10000001 0x34 0x10000002 0x56 0x10000003 0x78 0x10000004 0x9A 0x10000005 0xBC 0x10000006 0xDE 0x10000007 0x67 Exercise 10: Convert the following program into machine code. 0xFC00000C start: ……………… 0xFC000010 loop: addi $t0,$t0,-1 0xFC000014 sw $t0, 4($t2) 0xFC000018 bne $t0, $t3, loop 0xFC00001C j start addi $t0,$t0,-1 -1 = 0xFFFF 0010 00/01 000/0 1000 /1111 1111 1111 1111 8 8 8 -1 Op(6) rs(5) rt(5) immediate(16) sw $t0, 4($t2) 1010 11/01 010/0 1000 /0000 0000 0000 0100 43 10 8 4 Op(6) rs(5) rt(5) immediate(16) bne $t0, $t3, loop target address = (immediate * 4) + address of the following instruction immediate = ( target address – address of the following instruction) / 4 = (FC000010 – FC00001C) / 4 = - C /4 = -3 1111 1111 1111 1101 8 Computer Architecture Discussion Or convert to bainary first = 1111 1100 0000 0000 0000 0000 0001 0000 – 1111 1100 0000 0000 0000 0000 0001 1100 = 1111 1100 0000 0000 0000 0000 0001 0000 + 0000 0011 1111 1111 1111 1111 1110 0100 = 1111 1111 1111 1111 1111 1111 1111 0100 / 4 srl by 2 = 11 1111 1111 1111 1111 1111 1111 1101 immediate is 16 only so immediate = 1111 1111 1111 1101 0001 01/01 000/0 1011 /1111 1111 1111 1101 5 8 11 -3 Op(6) rs(5) rt(5) immediate(16) j start target address = last 4 bits of PC : (immediate * 4) immediate = first 28 bits from target address / 4 = C00000C / 4 = 3000003 = 1100 0000 0000 0000 0000 0000 1100 / 4 srl by 2 = 11 0000 0000 0000 0000 0000 0011 0000 10/11 0000 0000 0000 0000 0000 0011 2 3000003 Op(6) immediate (26) Exercise 11: (2.38 from book) Explain why an assembler might have problems directly implementing the branch instruction in the following code sequence: here: beq $s0, $s2, there … there: add $s0, $s0, $s0 Show how the assembler might rewrite this code sequence to solve these problems. 9 Computer Architecture Discussion The problem is that we are using PC-relative addressing, so if that address is too far away, we won’t be able to use 16 bits to describe where it is relative to the PC. If there refers to a location further than 128 KB from the PC, the solution would be: here: bne $s0, $s2, skip j there skip: … there: add $s0, $s0, $s0 If there refers to a location further than 256 MB from the PC, the solution would be: here: bne $s0, $s2, skip lui $ra, there(upper) ori $ra, $ra there(lower) jr $ra skip: … there: add $s0, $s0, $s0 Exercise 12: Suppose that you have already written a MIPS function with the following signature: int sum(int A[], int first, int last). This function calculates the sum of the elements of A starting with element first and ending with element last. Write a fragment of MIPS assembly language which calls this function and uses it to calculate the average of all values in A. You may assume that the size of the array A is N, the base address of A in $a0. Average: add $a1, $zero, $zero # index of first element addi $a2, $zero, N addi $a2, $a2, -1 # index of last element is N-1 jal sum add $t0, $zero, $v0 # Save the return value in $t0 addi $t1, $zero, N # Load size of array into $t1 div $t2, $t0, $t1 # This form of div is provided as a pseudoinstruction. 10 Computer Architecture Discussion Exercise 13: Below is a recursive version of the function BitCount. This function counts the number of bits that are set to 1 in an integer. Your task is to translate this function into MIPS assembly code. The parameter x is passed to your function in register $a0. Your function should place the return value in register $v0. int BitCount(unsigned x) { int bit; if (x == 0) return 0; bit = x & 0x1; return bit + BitCount(x >> 1);
}
BitCount:
addi $sp, $sp, -8
sw $s0, 4($sp)
sw $ra, 0($sp)
bne $a0, $0, else
add $v0, $0, $0
addi $sp, $sp, 8
jr $ra
else:
andi $s0, $a0, 1
srl $a0, $a0, 1
jal BitCount
add $v0, $v0, $s0
lw $ra, 0($sp)
lw $s0, 4($sp)
addi $sp, $sp, 8
jr $ra
11 Computer Architecture Discussion
Extra exercises
Exercise 14: (2.29 from book)
Add comments to the following MIPS code describe in one sentence what it computes. Assume
that $a0 and $a1 are used for the input and both initially contain the integers a and b,
respectively. Assume that $v0 used for the output.
add $t0, $zero, $zero # initialize running sum $t0 = 0
loop: beq $a1, $zero, finish # finished when $a1 is 0
add $t0, $t0, $a0 #compute running sum of $a0
sub $a1, $a1, 1 # compute this $a1 times
j loop
finish: addi $t0, $t0, 100 # add 100 to a * b
add $v0, $t0, $zero # return a * b + 100
t0=0
while(a1 != 0){
t0 = t0 + a0;
a1 = a1 – 1;
}
t0 = t0 + 100;
v0 = t0;
The program computes a * b + 100.
Exercise 15: (2.34 from book)
The following program tries to copy words from the address in register $a0 to the address in
register $a1, counting the number of words copied in register $v0. The program stops copying
when it finds a word equal to 0. You do not have to preserve the contents of registers %v1, $a0
and $a1. This terminating word should be copied but not counted.
addi $v0, $zero, 0 # Initialize count
loop:lw, $v1, 0($a0) # Read next word from source
sw $v1, 0($a1) # Write to destination
addi $a0, $a0, 4 # Advance pointer to next source
addi $a1, $a1, 4 # Advance pointer to next destination
beq $v1, $zero, loop # Loop if word copied != zero
There are multiple bugs in this MIPS program; fix them and turn in a bug-free version.
Bug 1: Count ($v0) is initialized to zero, not –1 to avoid counting zero word.
Bug 2: Count ($v0) is not incremented.
Bug 3: Loops if word copied is equal to zero rather than not equal.
12 Computer Architecture Discussion
Bug-free version:
addi $v0, $zero, -1 # Initialize to avoid counting zero word
loop:lw, $v1, 0($a0) # Read next word from source
addi $v0, $v0, 1 # Increment count words copied
sw $v1, 0($a1) # Write to destination
addi $a0, $a0, 4 # Advance pointer to next source
addi $a1, $a1, 4 # Advance pointer to next destination
bne $v1, $zero, loop # Loop if word copied != zero
Exercise 16:
Convert the following C fragment to equivalent MIPS assembly language. Assume that the variables a,
b, c, d, i and x are assigened to registers $t1, $t2, $t3, $t4, $s0 and $s1 respectively. Assume
that the base address of the array A and B is in register $a0 and $a1 respectively.
a) if ((a
bne $t3, $0, not #if (c!=0) go to not
addi $t4, $0, 1
not:
b) if (a > 0)
b = a + 10;
else
b = a – 10;
slt $t0, $0, $t1 # if $0 < $t1 then $t0 = 1, else $t0 = 0 beq $t0, $0, else # if $t0 == $0 then branch to else addi $t2, $t1, 10 j exit else: addi $t2, $t1, -10 exit: c) A[x+3] = B[x+2] | 0x10 addi $t0, $s1, 2 # $t0 = x+2 sll $t0, $t0, 2 # $t0 = (x+2)*4 add $t1,$a1,$t0 # $t1 = (base address of B + (x + 2)) lw $t2,0($t1) # $t2 = B[x+2] ori $t3,$t2,0x10 # $t3 = B[x+2] | 0x10 addi $t4,$s0,3 # $t4 = x+3 sll $t4, $t4, 2 # $t4 = (x+3)*4 add $t5,$a0,$t4 # $t5 = (base address of A + (x + 3)) sw $t3,0($t5) # A[x+3] = $t3 13 Computer Architecture Discussion d) for(int i=0; i<5; i++){ a += b; } add $s0, $zero, $zero Loop: stli $t0, $s0, 5 beq $t0, $zero, exit add $t1, $t1, $t2 addi $s0, $s0, 1 j Loop exit: Exercise 17: Convert this high level language code into MIPS code. Do not forget to write MIPS code for the abs(x) procedure. (i saved in $s0, the address of a in $s1, y in $s2) i = 3; y = y + abs(a[i]); addi $s0,$zero,3 sll $t0,$s0,2 add $t1,$t0,$s1 lw $a0,0($t1) jal abs add $s2,$s2,$v0 j exitall abs: slt $t0,$a0,$zero bne $t0,$zero,else add $v0,$a0,$zero jr $ra else: sub $v0,$zero,$a0 jr $ra exitall: Exercise 18: For each of the following, write the shortest sequence of MIPS assembly instructions to perform the specified operation. (Hint: 12345678 = 188 x 2 16 + 24910 31415924 = 479 x 2 16 + 24180) a) $v0 = 12345678 lui $v0, 188 ori $v0, $vo, 24910 14 Computer Architecture Discussion b) if ( $t0 < 12345678 ) go to address less lui $t1, 188 ori $t1, $t1, 24910 slt $t2, $t0, $t1 bne $t2, $zero, less c) t1 = 12345678 + 31415924 lui $t0, 188 ori $t0, $t0, 24910 lui $t2, 479 ori $t2, $t2, 24180 add $t1, $t0, $t2 Exercise 19: Given the register values in two's complement representation $s1 = 0000 0000 0000 0000 0000 0000 0000 0101 = 5 $s2 = 0000 0000 0000 0000 0000 0000 0000 0011 = 3 $s3 = 1111 1111 1111 1111 1111 1111 1111 1100 = -4 What are the values of registers $s1 through $s4 after executing the following MIPS instructions: slt $s1, $s1, $s2 Result: $s1 = 0 slt $s2, $s1, $s3 Result: $s2 = 0 sltu $s3, $s1, $s2 Result: $s3 = 0 sltu $s4, $s1, $s3 Result: $s4 = 1 sltu compares the unsigned values in the registers, so in the last instruction $s1 less than $s3 because $s3 has big unsigned value, so $s4 = 1. Best Wishes