Concept explainers
The following table represents the addresses and contents (using hexadecimal notation) of some cells in a machine’s main memory. Starting with this memory arrangement, follow the sequence of instructions and record the final contents of each of these memory cells:
Address | Contents |
0x00 | 0xAB |
0x01 | 0x53 |
0x02 | 0xD6 |
0x03 | 0x02 |
Step 1 Move the contents of the cell whose address is 0x03 to the cell at address 0x00.
Step 2 Move the value 0x01 into the cell at address 0x02.
Step 3 Move the value stored at address 0x01 into the cell at address 0x03.
Want to see the full answer?
Check out a sample textbook solutionChapter 1 Solutions
Computer Science: An Overview (12th Edition)
Additional Engineering Textbook Solutions
Software Engineering (10th Edition)
Java: An Introduction to Problem Solving and Programming (7th Edition)
Digital Fundamentals (11th Edition)
Starting Out with Java: Early Objects (6th Edition)
Concepts of Programming Languages (11th Edition)
Starting Out with Programming Logic and Design (4th Edition)
- Most Intel CPUs use the __________, in which each memory address is represented by two integers.arrow_forwardIn the ________, memory addresses consist of a single integer.arrow_forwardThe contents of memory location B0000, are FF 16, and those at B0001 6 are 0016. What is the data word stored at the address B0000? Is the word aligned or misaligned?arrow_forward
- A computer has a 256K word addressable memory Module with 16 bits per word. The instruction set consists of 166 different instructions. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. Based on the above, answer the following questions: a. How many bits are there in the main memory? (Represent it in power of 2) b. How many bits are needed for the opcode? c. How many bits are left for the address part of the instruction? d. How many additional instructions can be added to the existing 166 without affecting the assigned size of the opcode part? Justify.arrow_forwardThe computer program is stored in the lowest 1kbyte block of memory. Give the start and end address of this blockarrow_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 of memory.Q.) What is the maximum allowable size for memory?arrow_forward
- Memory 12200 12201 12202 12203 12204 Content %D AA EE FF 22 What result is produced in the destination operand by execution the following instruction? a- LEA SI[DI+Bx+5] b- LDS SI.[200]arrow_forwardThe ALU unit of a microprocessor manipulate logically or arithmetically on array of Bits. Below are two set of Bits equations placed in the ALU sequentially: Calculate 11100011 AND 11001110 and Calculate 01010111 OR 11001100 By your reasoning and application of principle of ALU solve this equations step by steparrow_forward8051 MICROCONTROLLER PROGRAMMING USING ASSEMBLY LANGUAGE, MOST OF THE CODES I GET FROM HERE USES 8086 MICROCONTROLLER, PLEASE DO USE 8051 ASSEMBLY LANGUAGE, I DO NOT NEED ONE WITH 8086. Given arbitrary 8-bit number, check whether it is bigger than signed decimal - 10, if it is bigger than -10, move #'Y' to R5, if not move #'N' to R5.arrow_forward
- All microprocessors have same number of address lines but different number of data lines. Select one: True Falsearrow_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 memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with four fields: an opcode field; a mode field to specify one of seven addressing modes; a register address field to specify one of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:Q.) How large is the opcode field?arrow_forward
- Systems ArchitectureComputer ScienceISBN:9781305080195Author:Stephen D. BurdPublisher:Cengage Learning