Concept explainers
a.
Instructions in Vole machine language:
Computer can understand only machine language, so the instruction must be given in vole machine language. It is an encoded language. The instructions are encoded by using total 16-bits that is also represented by hexadecimal digits. Instructions in vole machine consist of op-code and operation field of the register. First 4 bits are called the op-code and operation field has last 12 bits.
b.
Instructions in Vole machine language:
Computer can understand only machine language, so the instruction must be given in vole machine language. It is an encoded language. The instructions are encoded by using total 16-bits that is also represented by hexadecimal digits. Instructions in vole machine consist of op-code and operation field of the register. First 4 bits are called the op-code and operation field has last 12 bits.
c.
Instructions in Vole machine language:
Computer can understand only machine language, so the instruction must be given in vole machine language. It is an encoded language. The instructions are encoded by using total 16-bits that is also represented by hexadecimal digits. Instructions in vole machine consist of op-code and operation field of the register. First 4 bits are called the op-code and operation field has last 12 bits.
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Computer Science: An Overview (13th Edition) (What's New in Computer Science)
- A large endian byte addressed memory system with eight distinct memory modules is included in a computer system. Each memory module has 134217728 cells and is 32 bits wide. a) What is the 32-bit memory address of cell 1048578 in module 3 if the memory uses high order interleaving? b) What is the 32-bit memory address of byte 1048575 inside module 2 if the memory uses high order interleaving? c) What is the 32-bit memory address of cell 511 inside module 1 if the high order interleaved memory utilises little endian storage order instead of big endian storage order?arrow_forwardAssume the PC has the address 0x021A when FETCH INSTRUCTION phase begins. After that first instruction executes, which one describes the second instruction to execute? a) LDR R4 with a value from memory b) LDR R5 with a value from memory c) ADD R4 to R5 and save result in R7 d)arrow_forwardComputer Science Please answer this question in assembly language with .asm extension. The code given in 99Heater.asm file is: ; ===== Heater and Thermostst on Port 03 ==========================; ===== 99Heater.asm ==============================================; ===== Heater and Thermostst on Port 03 ========================== MOV AL,0 ; Code to turn the heater off OUT 03 ; Send code to the heater IN 03 ; Input from Port 03 AND AL,1 ; Mask off left seven bits JZ Cold ; If the result is zero, turn the heater on HALT ; Quit Cold: MOV AL,80 ; Code to turn the heater on OUT 03 ; Send code to the heater END; ================================================================= Fix the program 99Heater.asm so that the temperature will stay at 21 ºC. Please solve the question in assembly language. I will definitely give you THUMBS UP.arrow_forward
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- A computer system contains a big endian byte addressable memory system with 8 separate memory modules. Each memory module is 32 bits wide and contains 134217728 cells. Cells within each memory module are numbered 0 through 134217727. a) If the memory employs high order interleaving, what is the 32-bit memory address of cell 1048578 within module 3? b) If the high order interleaved memory uses little-endian instead of big-endian storage order, what is the 32-bit memory address of cell 511 within module 1? c) If the memory employs low order interleaving, what is the 32-bit memory address of byte 1048575 within module 2? Bytes are numbered starting from 0. Express the address in hex. d) What is the main advantage of using low-order interleaved memory compared to using high-order interleaved memory?arrow_forwardA digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word ofmemory.a. How many bits are needed for the opcode?b. How many bits are left for the address part of the instruction?c. What is the maximum allowable size for memory?d. What is the largest unsigned binary number that can be accommodated in one word of memory?arrow_forwardThe first two bytes of a 4M x 16 main memory have the following Hex values: Byte 0 is FE Byte 1 is 04 If these bytes hold a 16-bit two's complement integer, what is its actual decimal value if: a) Memory is big endian? b) Memory is little endian? 2. The memory unit of a computer has 2M Words of 32 bits (or 4 bytes) each. The computer has an instruction format with 4 fields: an opcode field; an addressing mode field to specify 1 of 6 addressing modes; a register address field to specify one of 7 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following: How large must the addressing mode field be? How large must the register field be? How large must the address field be? How large is the opcode field?arrow_forward
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- Systems ArchitectureComputer ScienceISBN:9781305080195Author:Stephen D. BurdPublisher:Cengage Learning