In the event that each page is 1 KB in size, what are the page numbers and offsets (please supply them as decimal integers) for the following address references? (please provide the proper page numbers and offsets, as well as the stages of the computation) a. 2562 b. 23563 c. 11500
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- In the event that each page is 1 KB in size, what are the page numbers and offsets (please supply them as decimal integers) for the following address references? (please provide the proper page numbers and offsets, as well as the stages of the computation)a. 2562 b. 23563 c. 11500Student 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 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 records. If a2 has the address of a student record, what is the location of the next record in the array / data area: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 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 records. If a2 has the address of a student record, if using indexed addressing mode, what is the index value X such that X(a2) addresses the course mark?
- 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 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 ?adres veri 01h 5x9 02h 5x8 8715683 b190100564 7156 08h 6715988 5x2 bis010054 71 09h 5x1 Write the asm code that will create the address and contents given in the table. (It is mandatory to use loop and indirect addressing.) b10100564 159 190100564-7I5
- Part A For each byte sequence listed, determine the Y86 instruction sequence it encodes. If there is some invalid byte in the sequence, show the instruction sequence up to that point and indicate where the invalid value occurs. For each sequence, the starting address, then a colon, and then the byte sequence are shown. 0x100: 30f3fcfffff40630008000000000000 0x100: 30f3fcfffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rrmovq %rsi,0×80A(%rcx) O0x115: 00 halt Ox100: 30f3fcffffffff irmovq $-4,%rbx Ox10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox114: 00 halt 0x100: 30f3fcfffffffff rrmovq $-8,%rbx Ox109: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) 0x200: a06f800c020000000000000030f30a00000000000000 0x113: 00 halt 0x100: 30f3fcffffffffff irmovq $-4,%rbx 0x200: a06f pushq %rsi 0x10a: 40630008000000000000 | rmmovq %rsi,0x800(%rbx) Ox202: 800c02000000000000 call proc Ox116: 00 halt 0x20b: 00 halt 0x20c: proc: Submit Request Answer 0x20c: 30f30a00000000000000 | irmovq $10,%rbx…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 identifierThere 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 records.Write an assembler subroutine GetMax that will scan the entire list and find the highest mark. If there is a tie, the first student in the list with the highest mark should be found. The subroutine should return the starting address of this record in address register a3.The memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.
- Which of these describes how an item in a doubly linked data structure is accessed? a.immediately accessed via a base address and offset b.accessed by starting at one end and following the links until the item is reached c.immediately accessed via an index d.accessed via a binary searchFind the register values:.datamyBytes BYTE 12h,34h,56h,78h.codemov ax,WORD PTR [myBytes] ; AX = ?mov ax,WORD PTR [myBytes+2] ; AX = ?mov eax,DWORD PTR myBytes ; EAX = ?.datamyBytes BYTE 10h,20h,30h,40hmyWords WORD 3 DUP(?),2000hmyString BYTE "ABCDE"12. Insert a directive in the given data that aligns myBytes to an even-numbered address.
