WRITE/STORE 10 NUMBERS IN TIIE MEMORY STARTING FROM ANY 16 BIT ADDRESS (F.G 2000 H). SHIFT THE ALTERNATE NUMBERS IO ANOTIIER MEMORY LOCATION IN TIIE REVERSED ORDER. REMAINING ALTERNATE NUMBERS TO BE SHIFTED TO SEPARATE MEMORY LOCATION IN THE SAME ORDER. WRITE A PROGRAM SEQUENCE.
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A: Below is the answer to above question. I hope this will be helpful for you.
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A: Given: Given that SS=2400, SP=8631H, AX=4FA6H, and DX=8C3FH. What the contents are of registers AH,…
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A: I have answered this question in step 2.
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A: Lets see the solution.
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A: Solution : From the given data of the Address and their value & Register and their value.…
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A: Input: d1 = 655 Output: 655 Input: d1 = 234 Output:234
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A: Answer : 1 Suppose 0002 is the base address of a segment. Now consider two memory locations 0002F…
Q: WRITE/STORE 10 NUMBERS IN TIIE MEMORY STARTING FROM ANY 16 BIT ADDRESS (F.G 2000 H). SHIFT THE…
A: The answer is
Q: Suppose a computer using direct-mapped cache has 232 bytes of byte-addressable main memory and a…
A: A cache is an equipment or programming segment that stores information so future solicitations for…
Q: WRITE/STORE 10 NUMBERS IN TIIE MEMORY STARTING FROM ANY 16 BIT ADDRESS (F.G 2000 H). SHIFT THE…
A: The answer is
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Q: WRITE/STORE 10 NUMBERS IN TIE MEMORY STARTING FROM ANY 16 BIT ADDRESS (F.G 2000 H). SHIFT THE…
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A: Please find the answer to the above question below:
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A: Actually, cache memory is a fast access memory.
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A: Given: 5. suppose a computer using fully associative cache has 224 bytes of byte-addressable main…
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A: It is answered in the below step.
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A:
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- Which of the following statements best describes this code that is attempting to print the value of a disk block. Note that syntax of queue has changed slightly from the assignment to simplify it. void interrupt_service_routine(){ int* result; void (*callback) (int*); queue_dequeue(&result, &callback); callback(result); void printInt(int* i){ printf("%d\n", *i); free(i); void read (int blockno){ malloc(sizeof(int)); queue_enqueue(result, printInt); disk_read(result, blockno); int* result It has an error : it might print before the read has completed. X O It makes incorrect use of function pointers. O It has a dangling pointer bug. OIt has an error: it never calls interrupt_service_routine. It has a memory leak bug. OIt is correct. X 0%2. Write a simple encryption/decryption program. Function encrypt takes a character pointer as a parameter and uses pointer-subscript notation to change the value in the address pointed to by adding 1 to it. Function decrypt takes a character pointer as a parameter and uses pointer notation to change the value in the address pointed to by subtracting 1 from it. Function main calls functions encrypt and decrypt and prints the encrypted string. // A simple encryption/decryption program #include "stdafx.h" #include using namespace std; // write the prototype for function encrypt // write the prototype function decrypt int main() { // create a string to encrypt char string[] = "this is a secret!"; cout « "Original string is : " « string « endl; encrypt(string); cout « "Encrypted string is : « string <« endl; decrypt(string); cout <« "Decrypted string is : «string<< endl;Assume that you are monitoring the rate at which the pointer in the clock algorithm (which indicates the candidate page for replacement) moves. What can you say about the system if you notice the following behavior: a. pointer is moving fast b. pointer is moving slow
- 156. Interrupt addresses are stored in a. array b. stack c. indexes d. queue1. Suppose an array with six rows and eight columns is stored in a row major order starting at address 20 (base 10). If each entry in the array requires only one memory cell, what is the address on the entry in the third row and fourth columns? What if each entry requires two memory cells?15. The only way to access data stored in heap memory is through pointer variables malloc () allocation tables а. b. С. d. a buffer memory
- Oeew s Sat DO D 4 Hene et e H P E Dei vi At D Metig eta O t TE Fie • REC In Net ASSIGNMENT 1. The table below presents a list of devices that are to be addressed in a certain memory space. They have been ordered in the manner in which S S they are to be addressed with the first component being placed on the upper end of memory, starting at address $000000. By considering the size each component, provide the start and end address using the appropriate hexadecimal value. l Device Description Device Name Amount of memory to address ROM Chip ROM 1 RAM 1 4KB RAM Chip 8KB ROM Chip ROM 2 4KB Peripheral PER 1 4 bytes Peripheral PER 2 2 bytes 2. Assume a very simple microprocessor with 12 address lines Let's assume we wish to implement all its memory space and we use 517x8 memory chigs. a. What is the size of the largest addressable memory? 1aa H Q O B CEOEUsing C Language In this function, b has the address of a memory buffer that is num_bytes long. The function should repeatedly copy the 16 byte pattern that pattern16 points at into the memory buffer until num_bytes have been written. If num_bytes is not a multple of 16, the final write of the 16 byte pattern should be truncated to finish filling the buffer. void memset16(void *b, int num_bytes, void *pattern16) For example if the 16 bytes that pattern16 points at is 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff, then memset(b, 20, pattern16) should write to the buffer pointed at by p the 20 bytes 00 11 22 33 44 55 66 77 88 99 aa bb cc dd ee ff 00 11 22 33. Use SSE instructions to improve efficiency. Here's pseudocode. x = SSE unaligned load from pattern16while (num_bytes >= 16) SSE unaligned store x to p advance p by 16 bytes decrement num_bytes by 16while (num_bytes > 0) store 1 byte from pattern16 to p advance p by 1 byte advance pattern16 by 1 byte…Select all true statements. Segmentation always uses 32-bit logical addresses. The memory management unit utilizes the segment component of the logical address to get the segment table start address and adds the offset to obtain the physical address. Segmentation uses segment and offset logical addresses. Maximum segment number limits segment length. Segmentation restricts process memory access to respective segments. Segments may be granted privileges.
- *Student information is being held in a data area, where each student record has the following format: The first nine bytes are the student number, held in ASCII The next byte is the course mark The next word is the section identifier 10009. There are well over three hundred such student records that have been loaded sequentially into memory starting at address $10000. The last record loaded is a dummy record with a section identifier of $FFFF, to show the end of the rècords. If a2 has the address of a student record, what is the location of the next record in the array / data area. Explain your answer.Weight 2. A BMI (body mass index) is roughly weight over height square (BMI : Height2 Write Assembly code to calculate average BMI of N students. Relevant data is stored in memory as shown in the example below. Address 10 has the total number of students (N) followed by N weights and heights. Store the computed average BMI at the address 70. For example, if N = 2, the memory contents will be as follows: 0 1 2 |3 | 4 |5 6 10 02 96 05 5A | 06 There are 2 records. Weight and height of the 1s person is 96 and 05 respectively. Similarly, 5A and 06 are the weight and height of the second person. Assume decimal points are ignored and the average will be a rounded integer value. All divisions are integer divisions as well. Please use the Instruction Set of the simulator discussed in class. (http://www.softwareforeducation.com/sms32v50/sms32v50_manual/245-IsetSummary.htm)Complete the following sequence of three MIPS instructions: mul28: $t0, 28 mul $to ir by dragging four items from the below list into the correct spaces. Your code should implement a procedure, using the standard conventions covered in class for passing parameters and returning results, that returns the product of its argument and 28. $ra li sb $a0 $v0 Ib move lui Give a sequence of two MIPS instructions that branches to the target address represented by the name "loop" if and only if the contents of $$0 are less than or equal to the contents of $s1. Do this by dragging two items from the below list into the correct spaces. slt $at, $s1, $s0 bne $at, $zero, loop bltz $at, loop slt $at, $s0, $s1 beq $at, $zero, loop