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3 years ago

Design an algorithm for a bounded-buffer monitor in which the buffers (portions) are embedded within the monitor itself."

Computers and Technology

1 answer:

sattari [20]3 years ago

5 0

Answer:

Required code is given below:

Explanation:

monitor bounded buffer {

int items[MAX ITEMS];

int numItems = 0;

condition full, empty;

void produce(int v)

{

while (numItems == MAX ITEMS) full.wait();

items[numItems++] = v;

empty.signal();

}

int consume()

{

int retVal;

while (numItems == 0) empty.wait();

retVal = items[--numItems];

full.signal();

return retVal;

}

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You are tasked to calculate a specific algebraic expansion, i.e. compute the value of f and g for the expression: ???? = (??????

melisa1 [442]

Answer:

.data

prompt: .asciiz "Enter 4 integers for A, B, C, D respectively:\n"

newLine: .asciiz "\n"

decimal: .asciiz "f_ten = "

binary: .asciiz "f_two = "

decimal2: .asciiz "g_ten = "

binary2: .asciiz "g_two = "

.text

main:

#display prompt

li $v0, 4

la $a0, prompt

syscall

#Read A input in $v0 and store it in $t0

li $v0, 5

syscall

move $t0, $v0

#Read B input in $v0 and store it in $t1

li $v0, 5

syscall

move $t1, $v0

#Read C input in $v0 and store it in $t2

li $v0, 5

syscall

move $t2, $v0

#Read D input in $v0 and store it in $t3

li $v0, 5

syscall

move $t3, $v0

#Finding A^4

#Loop (AxA)

li $t6, 0

L1:

bge $t6, $t0, quit

add $s1, $s1, $t0 # A=S+A => $s1= A^2

addi $t6, $t6, 1 # i=i+1

j L1

quit:

#Loop (A^2 x A^2)

li $t6, 0

L1A:

bge $t6, $s1, quit1A

add $s5, $s5, $s1

addi $t6,$t6, 1

j L1A

#End of Finding A^4

#Finding 4xA^3

quit1A:

#Loop (4xB)

li $t6, 0

L2:

bge $t6, 4, quit2

add $s2, $s2, $t1

addi $t6, $t6, 1

j L2

quit2:

#Loop (BxB)

li $t6 , 0

L2A:

bge $t6, $t1, quit2A #loop2

add $s6, $s6, $t1 #add

addi $t6, $t6, 1 #add immediate

j L2A #loop2

quit2A: # perform proper program termination using syscall for exit

#Loop (BxB)

li $t6 , 0 #load immediate

L2AA:

bge $t6, $s2, quit2AA #loop2

add $t7, $t7, $s6 #add

addi $t6, $t6, 1 #add immediate

j L2AA #loop2

#End ofFinding 4xA^3

#Finding 3xC^2

quit2AA: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3:

bge $t6, $t2, quit3 #loop3

add $s3, $s3, $t2 #add

addi $t6,$t6, 1 #add immediate

j L3 #loop3

quit3: # perform proper program termination using syscall for exit

#3 Loop (3 x (C x C)) FOR S3

li $t6 , 0 #load immediate

L3A:

bge $t6, 3, quit3A #loop3

add $s0, $s0, $s3 #add

addi $t6,$t6, 1 #add immediate

j L3A #loop3

#End of Finding 3xC^2

#Finding 2xD

quit3A: # perform proper program termination using syscall for exit

#4 Loop (2 x D) FOR S4

li $t6 , 0

L4:

bge $t6, 2, quit4 #loop4

add $s4, $s4, $t3 #add

addi $t6, $t6, 1 #add immediate

j L4 #Loop4

#End of Finding 2xD

#Finding AxB^2

quit4:

li $t6, 0

li $s1, 0

L5:

bge $t6, $t1, quit5

add $s1, $s1, $t1

addi $t6, $t6, 1

j L5

quit5:

li $t6, 0

li $s2, 0

L6:

bge $t6, $t0, quit6

add $s2, $s2, $s1

addi $t6, $t6, 1

j L6

#End of Finding AxB^2

#Finding C^2XD^3

quit6: #finds C^2

li $t6, 0

li $s1, 0

L7:

bge $t6, $t2, quit7

add $s1, $s1, $t2

addi $t6, $t6, 1

j L7

quit7: #finds D^2

li $t6, 0

li $s6, 0

L8:

bge $t6, $t3, quit8

add $s6, $s6, $t3

addi $t6, $t6, 1

j L8

quit8: #finds D^3

li $t6, 0

li $s7, 0

L9:

bge $t6, $t3, quit9

add $s7, $s7, $s6

addi $t6, $t6, 1

j L9

quit9: #finds C^2XD^3

li $t6, 0

li $s3, 0

L10:

bge $t6, $s1, end

add $s3, $s3, $s7

addi $t6, $t6, 1

j L10

#End of Finding C^2XD^3

end: # perform proper program termination using syscall for exit

#f is $t8

li $t8 , 0

sub $t8, $s5, $t7 # addition

add $t8, $t8, $s0 # subract

sub $t8,$t8, $s4 # subract

#g is $t9

li $t9 , 0

add $t9, $s2, $s3 # addition

#Display

#1st equation

li $v0,4 # display the answer string with syscall having $v0=4

la $a0, decimal # Gives answer in decimal value