Draw a picture illustrating the contents of memory, given the following data declarations: (You need to draw a diagram showing the memory addresses and their contents.) Assume that your data segment starts at 0x1000 in memory. Your answer should fill the table below and explain why.
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![Draw a picture illustrating the contents of memory, given the following data
declarations:
(You need to draw a diagram showing the memory addresses and their
contents.)
Assume that your data segment starts at 0x1000 in memory. Your answer should fill
the table below and explain why.
Addres Hexadecimal Hexadecimal
S
Value
Value
0x1000
Ox
|0x
Ox
Ox
0x
Name: asciiz "Jones"
Age:
byte 48
Numbers: word 11, 20
Letter1: asciiz "Car"
Letter2: .byte 25
Hexadecimal
Value
Hexadecimal
Value](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2b9a25ce-7950-429e-b571-3c0f034f6fd8%2Fd70d65c0-f318-4360-9b94-3e9ae5965c83%2Fd6rh41n_processed.jpeg&w=3840&q=75)
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- Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Oxppppdddd Page | Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 pppp: page number dddd: page offset 1 1 Oxd000 65536 2 2 Oxe000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0000ee00 ? What is the physical address for Ox00020001 ? What is the logical address for Oxe0001234 ? What is the logical address for Oxc0004268 ?3. We've seen variable types that store characters, integers and floating point numbers. In a very similar fashion, addresses can be are stored. A variable that stores an address value is called a pointer variable, or simply a pointer. Let's take a look at the following example. As you may have noticed, we did not directly assign anything to num, c, count, and salary. Instead, we used the pointers to assign values to where those pointers were pointing. What does this program do? Can you make changes to assign anything to num, c, count, and salary and print them using pointers numptr, cptr, countptr, and salaryptr. #include using namespace std; int main() { int num; char c; int count; float salary; // Declaring bunch of pointers that point to nothing (point to null) // declare a pointer variable to an integer // declare a pointer variable to a character // declare a pointer variable to an integer int char *numptr; *cptr; *countptr; *salaryptr; // declare a pointer variable to a float…Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Охрppdddd Page | Frame Frame Size (hex) Size (dec) 2 Охс000 Ox10000 65536 pppp: page number dddd: page offset 1 1 Оxd000 Ox10000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0002ffff What is the physical address for Ox0000abcd ? What is the logical address for Oxf000000f ? What is the logical address for Oxc000bbcc ?
- Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Page Table Physical Memory Physical Address (starting) Oxppppdddd Page Frame Frame Size (hex) Size (dec) Ox10000 Ox10000 2 Охс000 65536 PPpp: page number dddd: page offset 1 1 Оxd000 65536 3 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x00011119 What is the physical address for 0x00000001 What is the logical address for Oxd0000001 ? What is the logical address for Oxc0000002 ?Assume variables have logical addresses with 16-bit page numbers and 16-bit offset using the memory configuration below. (Note that each hexidecimal is 4 bits long and Ox means hexadecimal radix) Logical Address Format Physical Memory Physical Address (starting) Page Table Size (dec) 65536 65536 Frame Size (hex) Oxppppdddd Page Frame 2 Ox10000 Ox10000 Охс000 pppp: page number dddd: page offset 1 1 Oxd000 2 Охе000 Ox10000 65536 3 Oxf000 Ox10000 65536 Translate the following addresses: What is the physical address for 0x0002ffff ? What is the physical address for 0x0000abcd ?ADD [R1], R2, [R3]; Here [R1] and [R3] indicate memory locations pointed by R1 and R3 register respectively. Here the operand field next to opcode will hold the result at the end. Assume that the machine code of this instruction is loaded at address 1020H of the main memory. Also assume that the contents of registers R1, R2 and R3 are 2001H, 2002H and 2003H respectively. Moreover, 1000H, 2000H and 3000H are saved at memory addresses 2001H, 2002H and 2003H respectively. a) Draw a schematic diagram of a CPU, show its important functional units required to process this instruction.
- The following byte sequence is the machine code of a program function compiled with the Y86- 64 instruction set. The memory address of the first byte is 0x300. Note that the byte sequence is written in hex-decimal form, i.e., each number/letter is one hex-decimal number representing 4 binary bits, and two numbers/letters represent one byte. 630030F3020000000000000030F11E000000000000007023030000000 00000601061316211761F0300000000000090- Please write out the assembly instructions (in Y86-64 instruction set) corresponding to the machine codes given by the above bytes sequence, and explain what this program function is computing. The machine has a little-endian byte ordering.Computer organization and assembly language Please help me with this. I have to write line by line what each line of codes does. CODE IS BELOW: .model small .386 .stack 100h .data msg1 db 13, 10, "Enter any number --> ", "$" msg2 db "Enter an operation +,- * or / --> ",13, 10, "$" msg3 db "The Operation is --> ", "$" msg4 db "The result is --> ", "$" By_base dd 21 by_10 dd 10 ; 32 bits variable with initial value = 10 sp_counter db 0 ; 8 bits variable with initial value of zero disp_number dd 0 ; 32 bits variable with initial value = 0 disp_number2 dd 0 disp_number3 dd 0 op_type db 0 last_key dd 0 ; 32 bits variable with initial value of zero remainder db 0 .code main proc mov ax,@data;set up datasegment movds,ax mov dx,offset msg1 call display_message callm_keyin calloperation mov dx,offset msg1 calldisplay_message callm_keyin cmpop_type, "+" jnz short skip_plus callop_plus skiP_plus: cmp op_type, "-" jnz short skip_minus callop_minus…Write a PEP/8 machine language simulator that simulates the PEP/8 computer and instruction set for only the following instructions Here are some specifics. 1. Use an array to represent the memory. 2. Use variables or arrays for the PEP8 registers. I recommend putting it all into a structure. I also recommend using an array of 16 bit values for A, X, PC, SP so that you can use the r bit from the instructions to point directly to A or X. You do NOT have to include the Status bits. 3. Use unions of structures to break up the registers and instructions into the correct bits (for example, use a structure that can be unioned to break up the 8 bit specifier into the following bit combos (4, 1, 3), (5, 3), (7, 1), (8). This will allow you to instantly extract the instruction, register and addressing modes from each instruction. Also use a union to break up the 16 bit operand into two 8 bit values. 4. Have the instructions be inputted either through the command line or better yet,…
- Consider a program consists of five segments: S0 = 600, S1 = 14 KB, S2= 100 KB, S3 =580 KB and S4 = 96 KB. Assume at that time, the available free space partitions of memory are 1200–1805, 50 – 150, 220-234, and 2500-3180.Find the following:a. Draw logical to physical maps and segment table?b. Allocate space for each segment in memory?c. Calculate the external fragmentation and the internal fragmentation?d. What are the addresses in physical memory for the following logical addresses: 0.580, (b) 1.17 (c) 2.66 (d) 3.82 (e) 4.20?c programming language The program below uses pointer arithmetic to determine the size of a 'char'variable. By using pointer arithmetic we can find out the value of 'cp' and thevalue of 'cp+1'. Since cp is a pointer, this addition involves pointer arithmetic:adding one to a pointer makes the pointer point to the next element of the sametype.For a pointer to a char, adding 1 really just means adding 1 to the address, butthis is only because each char is 1 byte.1. Compile and run the program and see what it does.2. Write some code that does pointer arithmetic with a pointer to an int anddetermine how big an int is.3. Same idea – figure out how big a double is, by using pointer arithmetic andprinting out the value of the pointer before and after adding 1.4. What should happen if you added 2 to the pointers from exercises 1through 3, instead of 1? Use your program to verify your answer.#include <stdio.h>int main( ){ char c = 'Z'; char *cp = &c; printf("cp is %p\n", cp);…Computer Science Write a C program that calls malloc() 1001 times allocating blocks of size 4 each. Use type char * to store the pointers returned by malloc(). When done, print the difference in the addresses of the first and last block allocated. This difference is the amount of memory malloc actually used to create 1000 blocks of size 4. It is a lot more than you would expect. Repeat this for other block sizes and make a table of your results. Do you see any pattern? Any guesses why?
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