Assume a computer that has 16-bit integers. Show how each of the following values would be stored sequentially in memory in Little Endian order, assuming each address holds one byte sure to extend each value to the appropriate number of bits. Please just enter the value. a. 0x1AB2 b. OxA Blank # 1 Blank # 2 Blank # 3 A/
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- Assume a computer that has 32-bit integers. Use a table to clearly show how each of the following values would be stored sequentially in memory, starting at address 0x100, assuming each address holds one byte. 0x123456BA 0xabcde -28 (decimal values, assume the machine uses 2's complement notation. 0xFEDC6. Assume a computer has 32-bit integers. Show how the value 0x0001122 would be stored sequentially in memory, starting at address 0x000, on both a big endian machine and a little endian machine, assuming that each address holds one byte. Address Big Endian Little Endian0x000 0x001 0x002 0x003Show how the following values would be stored by byteaddressable machines with 32-bit words, using little endian and then big endian format. Assume that each value starts at address 10 . Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations.Q.) 0x0000058A
- Consider memory storage of a 32-bit word stored at memory word 34 in a byte addressable memory. (a) What is the byte address of memory word 34? (b) What are the byte addresses that memory word 34 spans? (c) Draw the number 0x3F526372 stored at word 342 in both big-endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.1. Draw a picture illustrating the contents of memory, given the following data declarations: You need to mark all the memory addresses. Assume that your data segment starts at 0x1000 in memory. Name: .asciiz "James Bond!"Age: .byte 24Numbers: .word 11, 22, 33Letter1: .asciiz "M"Consider the following image that represents part of the memory of a 16-bit address space that has an addressability of 2 bytes (like LC-3): A memory location can store an address. We call that memory location's contents a "pointer" since it's an address that "points" to another memory location. G.) Interpret the contents at address 0x0C0B as a pointer. (Enter hex like the following example: 0x2A3F) H.) What are the contents of the memory location that the pointer above is pointing to? (Enter hex like the following example: 0x2A3F) Another reference : LC-3 Opcodes in Hex ADD 0x1 JMP 0xC LDR 0x6
- Question Show how the following values would be stored bybyte-addressable machines with 32- bit words, using little endianand then big endian format. Assume each value starts at address301816. Draw a diagram of memory for each, placing the appropriatevalues in the correct (and labeled) memory locations. a. 56789ABC16 b. 2014111910 The memory unit of a computer has256K words of 32 bits each. The computer has an instruction formatwith 4 fields: an opcode field; a mode field to specify 1 of 7addressing modes; a register address field to specify one of 16registers; and a memory address field. Assume an instruction is 32bits long. Answer the following: a. How large must the mode field be? b. How large must the register field be? c. How large must theaddress field be? d. How large is the opcode field? 4. In a computerinstruction format, the instruction length is 12 bits and the sizeof an address field is 4 bits. Is it possible to have: 13 2-address instructions 45 1-address instructions…Suppose a computer system uses 16-bit addresses for both its virtual and physical addresses. In addition, assume each page (and frame) has size 256 bytes. Could you hand draw the page table, if possible a) Suppose that the OS uses a two-level page table. Draw the page table. (Assume that frames 7 through 221 are free, so you can allocate space for the page table there.) In addition, suppose that the page-table directory storage comprises a whole number of consecutive full frames. (For examples: if the directory entry is 2 bytes, the entry’s storage comprises 1 frame; if the directory entry is 260 bytes, the entry’s storage comprises 2 consecutive frames.) b) What is the size of the two-level page tableAssume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to = 8 the third is equal to = 16 help me find the last word
- Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4. Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at?Show how the following values would be stored by byte addressable machines with 32- bit words, using the little endian and then big endian. Assume each value starts at address 10 base 16. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. a. 456789A116 b. 0000058A16Show how the following values would be stored by machines with 32-bit words, using little endian and then big endian format. Assume each value starts at address 1016. Draw a diagram of memory for each, placing the appropriate values in the correct (and labeled) memory locations. 1234567816 ABCDEF1216 8765432116