a) A small amount of memory (RAM) in a computer-based device. For example, 64MB in a smartphone is considered low memory.
b) Address space is the amount of memory allocated for all possible addresses for a computational entity, such as a device, a file, a server, or a networked computer. Address space may refer to a range of either physical or virtual addresses accessible to a processor or reserved for a process.
c) Stored-program concept, Storage of instructions in computer memory to enable it to perform a variety of tasks in sequence or intermittently. The idea was introduced in the late 1940s by John von Neumann, who proposed that a program be electronically stored in binary-number format in a memory device so that instructions could be modified by the computer as determined by intermediate computational results.
d) In computer science, self-modifying code is code that alters its own instructions while it is executing – usually to reduce the instruction path length and improve performance or simply to reduce otherwise repetitively similar code, thus simplifying maintenance.
e)
The central processing unit (CPU) is the unit which performs most of the processing inside a computer. To control instructions and data flow to and from other parts of the computer, the CPU relies heavily on a chipset, which is a group of microchips located on the motherboard.
The CPU has two components:
To function properly, the CPU relies on the system clock, memory, secondary storage, and data and address buses.
This term is also known as a central processor, microprocessor or chip.
f) The instruction decoder of a processor is a combinatorial circuit sometimes in the form of a read-only memory, sometimes in the form of an ordinary combinatorial circuit. Its purpose is to translate an instruction code into the address in the micro memory where the micro code for the instruction starts.
2-1. Outline briefly what is meant by the following terms: (a) low memory (b) address space...
QUESTION 16 In the ARM Cortex MO core which register holds the memory address of the next instuction to be loaded from memory and usually increments by 2 when each instruction is executed? O a. Address register b.Instruction register C. Register bank d. Arithmetic and logic unit (ALU) Program counter ■ f. Instruction decoder ■ g. Control unit Z e.
How many address lines are required for a memory containing a total at (a) 256 kB (b) 2 MB (c) 4 GB Fill in the blanks with one of IR, ALU, PC, ACC, MEMORY, FETCH PHASE or EXECUTE PHASE: (a) Data is stored in the __________ (b) Programs are stored in the ___________ (c) To be executed, an instruction is loaded into the ____________ (d) The address of an instruction to be executed is held in ______________ (e) Instructions are...
-The memory address stored in 'b' is copied into a: D Question 4 1 pts A pointer provides an indirect means of accessing the value of a particular data item. ○True False D Question 5 2 pts What is the output of the following program
Problem #1 (25 points) Address Space, Memory Consider a hypothetical 18-bit processor called HYP18 with all registers, including PC and SP, being 18 bits long. The smallest addressable unit in memory is an 8-bit byte. A. (4 points) What is the size of HYP18's address space in bytes and KB? How many address lines does HYP18 require? Address space: Bytes Address space: KB (KiloBytes). Address bus lines: B. (6 points) Assume that first quarter of the address space is dedicated...
Question 3: ARM Processor a) What is the number of bits in a general-purpose register (e.g., R1) of the ARM Cortex-M4 processor (CPU)? b) What is the number of bits in a memory address for the ARM processor architecture? c) What is the number of bits in an assembly instruction for the ARM Thumb-2 instruction set? d) Consider the memory map used with the TM4C123 microcontroller shown below. If the stack is in data memory, what is the initial address...
Solve the following problems clearly - assembly - computer
organization and architecture- william stallings
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Let the address stored in the program counter be designated by the symbol X1. The instruction stored in X1 has an address part (operand reference) X2. The operand needed to execute the instruction is stored in the memory word with address X3. An index register contains the value X4. What is the relationship between these various quantities if the addressing mode of the instruction...
please answer all
questions!
I. Briefly explain what is meant by each of the following terms as it relates to this experiment. (a) Precipitation reaction (b) Spectator ion (c) The designation (ag) following the formula of a compound (d) Decomposition reaction (e) Coefficients II. (a) Write the complete ionic equation for the double displacement reaction that occurs when aqueous solutions of barium nitrate Ba(NO3)2 and sodium chromate Na2Cr04 are mixed. (b) Name the spectator ions in this chemical reaction. (c)...
2. Suppose that a 16M X 16 main memory is built using 512K X 8 RAM chips and memory is word addressable. a) How many RAM chips are necessary? ______ b) How many RAM chips are needed for each memory word? _______ c) How many address bits are needed for each RAM chip? _______ d) How many address bits are needed for all memory? _______ A digital computer has a memory unit with 24 bits per word. The instruction set...
Please show working. Thanks
What should be the value at memory X3012 if the following program is executed (based on the initial memory status)? Program ORIG x3000 The first instruction is at x3000 LD R1, #2 LD R2,#0 ST R1, #15 ST R2,#15 Initial Memory Status: Address: Content x3000: x2202 x3001: x2400 x3002: x320F x3003: x340F Starting from x3004, all of them are x0000 Select one a. x0000 b. x320F x340F d. x2202 e. X2400
What should be the value...
(2+2+1 5 points) Problem 10.1: simple cpu machine code = The following program has been written for the simple central processing unit introduced in class. The table below shows the initial content of the 16 memory cells. The first column denotes the memory address. Machine Code Assembly Code Description 001 1 0001 0 010 0 1111 1 001 1 0000 3 101 1 0100 110 1 0110 4 111 1 0000 5 6 001 0 001 1 100 1 0001...