Complete the following code. The goal is to implement the producer-consumer problem. You are expected to extend the provided C code to synchronize the thread operations consumer() and producer() such that an underflow and overflow of the queue is prevented. You are not allowed to change the code for implementing the queue operations, that is the code between lines 25 and 126 as shown in the Figure below. You must complete the missing parts between lines 226-261 as shown in the screenshot. ܕ : ܘܢܬܬܢ ܪ ܪ ܪ ܪ . ܝ . . ܫ ܕ13#include 14 #include 15 #include 16 #include 17 #define MAX_LENGTH_CAP 100 You can modify MAX_LENGTH_CAP18#define INIT -12719 #define UNDERFLOW (0x80 + 0x02)20 #define OVERFLOW 0x80+ 0x0121 #define BADPTR (0x80 + 0x03)22 #define CONSUMER_TERMINATION_PROBABILITY 4023 #define PRODUCER_TERMINATION PROBABILITY 302425 //262728 sem_t *FULLSPOTS;29 sem_t *EMPTYSPOTS;You can modify these two constantsfor test and experimentation//sem_t ACCESS; // use a mutex instead of a binary semaphorepthread_mutex_t ACCESS = PTHREAD_MUTEX_INITIALIZER;30 struct timespec remaining;31 struct timespec requested_time;32 > struct Queue{---383940> void enqueue (struct Queue *q, char x) -65 > char dequeue(struct Queue *q) -97 > void build(struct Queue **q, unsigned int length) ---125126 //127128 void *producer ( void *qptr ) ---179180 void *consumer( void *qptr) ...225226 void closeup (void);227 int main(int argc, const char * argv[])--308309> void closeup (void) ...LOCKEDUNLOCKEDDon't modify this code segmentTOTAL f225226 void closeup (void);227 int main(int argc, const char * argv[])228 {229230231232233234235236237238239240241242243244245246247248249250251252253254255256257258259260261262263pthread_t thread [6]; // declares the identifier for the thread we will create.pthread_attr_t attr;int iret [6];int q_test_size = 5;struct Queue *qptr = NULL;build (&qptr, q_test_size);(&attr);closeup();FULLSPOTS = (sem_t*) sem_open ("/FULLSPOTS", O_CREAT 10_EXCL, 0644,if (}perror("sem_open");exit (EXIT_FAILURE);EMPTYSPOTS = sem_open("/EMPTYSPOTS", O_CREATIO_EXCL, 0644,ifperror("sem_open");exit (EXIT_FAILURE);}int x=0;for(; x<5; x++) {iret [x] = pthread_create(&thread [x], &attr, producer, qptr); // creat 5 producersif (!iret [x]) // check for successprintf("PRODUCER_%d was created successfully.", x);}iret [x] = pthread_create(&thread [x], &attr, consumer, qptr); // ctreate one consumerif (!iret [x])printf("CONSUMER_%d was created successfully.", x);for(int x=0; x<6; x++) // Wait for all threads; verify the for-loopcloseup ();if (qptr->data) free(qptr->data);if(qptr) free (qptr);264265266267268269> void closeup (void) --printf("\n");return 0;& mainⓇ⁰ A⁰Figure 2 You must complete the missing parts. between lines 226 - 261 as shown in this screenshot> TOTAL

The missing parts of the code between lines 226 and 261 deal with thread synchronization for the…

#include #include #include #include #define MAX_LENGTH_CAP 100 You can modify MAX_LENGTH_CAP #define INIT -127 #define UNDERFLOW (0x80 + 0x02) #define OVERFLOW 0x80+ 0x01 #define BADPTR (0x80+ 0x03) #define CONSUMER_TERMINATION PROBABILITY 40 #define PRODUCER_TERMINATION_PROBABILITY 30 // =========== //sem_t ACCESS; // use a mutex instead of a binary semaphore pthread_mutex_t ACCESS = PTHREAD_MUTEX_INITIALIZER; sem_t *FULLSPOTS; sem_t *EMPTYSPOTS; struct timespec remaining; struct timespec requested_time; struct Queue{- void enqueue(struct Queue *q, char x)- char dequeue (struct Queue *q) - wvoid build(struct Queue **q, unsigned int length) - 1/ ========== void *producer(void *qptr ) --- You can modify these two constants for test and experimentation woid *consumer (void *qptr) --- void closeup (void); int main(int argc, const char * argv[])- ======== LOCKED ========== UNLOCKED ================ Don't modify this code segment

#include #include #include #include #define MAX_LENGTH_CAP 100 You can modify MAX_LENGTH_CAP #define INIT -127 #define UNDERFLOW (0x80 + 0x02) #define OVERFLOW 0x80+ 0x01 #define BADPTR (0x80+ 0x03) #define CONSUMER_TERMINATION PROBABILITY 40 #define PRODUCER_TERMINATION_PROBABILITY 30 // =========== //sem_t ACCESS; // use a mutex instead of a binary semaphore pthread_mutex_t ACCESS = PTHREAD_MUTEX_INITIALIZER; sem_t FULLSPOTS; sem_t EMPTYSPOTS; struct timespec remaining; struct timespec requested_time; struct Queue{- void enqueue(struct Queue q, char x)- char dequeue (struct Queue q) - wvoid build(struct Queue **q, unsigned int length) - 1/ ========== void producer(void qptr ) --- You can modify these two constants for test and experimentation woid consumer (void qptr) --- void closeup (void); int main(int argc, const char * argv[])- ======== LOCKED ========== UNLOCKED ================ Don't modify this code segment

C++ for Engineers and Scientists

4th Edition

ISBN:9781133187844

Author:Bronson, Gary J.

Publisher:Bronson, Gary J.

Chapter10: Pointers

Section10.1: Addresses And Pointers

Problem 5E

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switch (c) { case 1: a = 10; break; case 2: a = 20; break; case 3: a = 30; break; default: a = 0; } Translate this piece of C codes into ILOC codes. Variable c and a should be stored in the activation record. Your translated codes start at address 1000 and grow to larger addresses. The address of a variable in the activation record is r_arp[@name], where r_arp is the register storing the activation record address and @name is the macro for the offset of the variable name. You may use labels instead of address number in jump/branch instructions.
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zy Section 1.3 - IT 140: Introduct zy Section 1.3 - IT 140: Introduct X O (158) SNHU IT140 Module Or x b Answered: Run the program G Many useful programs allow + Ô https://learn.zybooks.com/zybook/SNHUIT140V3/chapter/1/section/3?content_resource_id=40688398 = zyBooks My library > IT 140: Introduction to Scripting v3 home > 1.3: Input/Output E zyBooks catalog ? Help/FAQ 8 Jose Roque CHALLENGE 1.3.4: Read user input numbers and perform a calculation. АCTIVITY Read two numbers from user input. Then, print the sum of those numbers. Hint -- Copy/paste the following code, then just type code where the questions marks are to finish the code. num1 = int(input ()) num2 = ? print(num1 + ?) Note: Our system automatically runs your program several times, trying different input values each time to ensure your program works for any values. Notes on how to solve this. See How to Use zyBooks for info on how our automated program grader works. 247772.2002516.qx3zqy7 1 2 ' Your solution goes here '' 1…
Home work1 Java Network programming You are required to program a client and a server application using TCP based socket. The client should send a message to the server of the following format Operation operand1 operand2 Where Operation is either add, subtract, multiply, or divide. Operand 1 and Operand2 are two integers. The server should distingues the required operation and perform it over the two operands. The result of carrying out this operation should be sent back to the client. The client should present the result to the user. The work have to be within a group of two. One student should focus on implementing the client side while the other should focus on the server. Both student should fully understand and be able to modify and explain both sides the client and the server. Rubrics are as follows 1- Well organized commented out code. (25%) 2- Correct implementation of the protocol described above. (25%) 3- Clear and easy to use interface with the user. (20%) 4- Fully running…
го, r10 Øx51dd4 ldrex ro, [r6] го, г5 Оx51а14 strex r1, r4, (r6] г1, #0 Øx51a04 Оx519fe: mov bl Оx51а00: Øx51a04: Оx51а08: cmp bne 0x51а0а: flag.compare_exchange_strong (...) Оx51а0с: Øx51a10: cmp bne Øx51a12: ro, #0 Øx519fe го, [r11] г8, г8, #1 ro, #1 го, [r11] г5, [r6] ro, #38527 ro, #152 г8, го Øx519fe Оx51а14: cmp bne ldr.w add.w Øx51a16: Øx51a18: Оx51а1с: sharedValue++ Øx51a20: adds Оx51а22: str.w flag.store (...) Øx51a26: str Оx51а28: movw Оx51а2с: movt Оx51а30: cmp ble Оx51а32:
DATA1 DW 12CH DATA2 DB 40 DATA3 DB 1,2 'DATA 4 DB '5' For this declared labels, starting with offset 0000h. which statement is true address 0002-40h address 0001=01h address 0005=05h address 00003D0CH all of them 6.
Explain what the following VHDL code does. [20 marks] EX_VAR : process (i_clk) variable v_Count : natural range 0 to 5 := 0; begin if rising_edge(i_clk) then v_Count := v_Count + 1; 20 21 22 23 24 25 26 r_Var_Copy1 <= v_Count; 27 if v_Count = 5 then r_Done <= '1'; v_Count := 0; else 28 %3D 29 30 31 r_Done <= '0'; end if; 32 33 34 35 r_Var_Copy2 <= v_Count; 36 end if; end process EXx_VAR; 37 38 39 40 o done <= r_Done; 41 42 end rtl;
The following code should use Timer A0 to trigger an ISR every 2ms. void main(void) { WDTCTL WOTPW | WDTHOLD; TASSEL_1 + ID_3 + MC_1 + TACLR; TAOCTL //Question 2c) //Question 2d) _enable_interrupt(); while(1); #pragma vector = TIMERO_A®_VECTOR interrupt void TIMERO_AO_ISR(void) { /* content not relevant */ } Figure Q3: Timer AO example code (a) What is the frequency of the ISR execution (i.e., how often is the ISR called per second)? (b) Based on the TAOCTL register configuration in the code, explain how Timer AO is configured. (c) Using an ACLK running at 32 kHz, calculate the correct Capture/Compare value and write the instruction to load this value into the appropriate register. (d) Write the instruction to enable Capture/Compare interrupts.