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ipn [44]
3 years ago
7

Read the introduction (paragraphs 1-3].

Computers and Technology
1 answer:
MArishka [77]3 years ago
3 0
Yes so drunk drivers are more wreck less
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After adding text to a shape, a user can make the text bold and change its color using options in the
lozanna [386]

Answer:

Top area of Word, the second segment.

Explanation:

To make it bold, press the "B" button and to change the colour press the "A" underlined with a colour (normally red).

6 0
3 years ago
There will be 10 numbers stored contiguously in the computer at location x 7000 . Write a complete LC-3 program, starting at loc
Artist 52 [7]

Answer:

The LC-3 (Little Computer 3) is an ISA definition for a 16-bit computer. Its architecture includes physical memory mapped I/O via a keyboard and display; TRAPs to the operating system for handling service calls; conditional branches on N, Z, and P condition codes; a subroutine call/return mechanism; a minimal set of operation instructions (ADD, AND, and NOT); and various addressing modes for loads and stores (direct, indirect, Base+offset, PC-relative, and an immediate mode for loading effective addresses). Programs written in LC-3 assembler execute out of a 65536 word memory space. All references to memory, from loading instructions to loading and storing register values, pass through the get Mem Adr() function. The hardware/software function of Project 5 is to translate virtual addresses to physical addresses in a restricted memory space. The following is the default, pass-through, MMU code for all memory references by the LC-3 simulator.

unsigned short int get Mem Adr(int va, int rwFlg)

{

unsigned short int pa;

// Warning: Use of system calls that can cause context switches may result in address translation failure

// You should only need to use gittid() once which has already been called for you below. No other syscalls

// are necessary.

TCB* tcb = get TCB();

int task RPT = tcb [gettid()].RPT;

pa = va;

// turn off virtual addressing for system RAM

if (va < 0x3000) return &memory[va];

return &memory[pa];

} // end get MemAdr

Simple OS, Tasks, and the LC-3 Simulator

We introduce into our simple-os a new task that is an lc3 Task. An lc3 Task is a running LC-3 simulator that executes an LC-3 program loaded into the LC-3 memory. The memory for the LC-3 simulator, however, is a single global array. This single global array for memory means that alllc3 Tasks created by the shell use the same memory for their programs. As all LC-3 programs start at address 0x3000 in LC-3, each task overwrites another tasks LC-3 program when the scheduler swaps task. The LC-3 simulator (lc3 Task) invokes the SWAP command every several LC-3 instruction cycles. This swap invocation means the scheduler is going to be swapping LC-3 tasks before the tasks actually complete execution so over writing another LC-3 task's memory in the LC-3 simulator is not a good thing.

You are going to implement virtual memory for the LC-3 simulator so up to 32 LC-3 tasks can be active in the LC-3 simulator memory without corrupting each others data. To implement the virtual memory, we have routed all accesses to LC-3 memory through a get Mem Adr function that is the MMU for the LC-3 simulator. In essence, we now have a single LC-3 simulator with a single unified global memory array yet we provide multi-tasking in the simulator for up to 32 LC-3 programs running in their own private address space using virtual memory.

We are implementing a two level page table for the virtual memory in this programming task. A two level table relies on referring to two page tables both indexed by separate page numbers to complete an address translation from a virtual to a physical address. The first table is referred to as the root page table or RPT for short. The root page table is a fixed static table that always resides in memory. There is exactly one RPT per LC-3 task. Always.

The memory layout for the LC=3 simulator including the system (kernel) area that is always resident and non-paged (i.e., no virtual address translation).

The two figures try to illustrate the situation. The lower figure below demonstrates the use of the two level page table. The RPT resident in non-virtual memory is first referenced to get the address of the second level user page table or (UPT) for short. The right figure in purple and green illustrates the memory layout more precisely. Anything below the address 0x3000 is considered non-virtual. The address space is not paged. The memory in the region 0x2400 through 0x3000 is reserved for the RPTs for up to thirty-two LC-3 tasks. These tables are again always present in memory and are not paged. Accessing any RPT does not require any type of address translation.

The addresses that reside above 0x3000 require an address translation. The memory area is in the virtual address space of the program. This virtual address space means that a UPT belonging to any given task is accessed using a virtual address. You must use the RPT in the system memory to keep track of the correct physical address for the UPT location. Once you have the physical address of the UPT you can complete the address translation by finding the data frame and combining it with the page offset to arrive at your final absolute physical address.

A Two-level page table for virtual memory management.

x7000 123F x7000 0042

x7001 6534 x7001 6534

x7002 300F x7002 300F

x7003 4005 after the program is run, memory x7003 4005

x7004 3F19

7 0
3 years ago
Read 2 more answers
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natita [175]
It’s the last one! The key to an effective persuasion is to know what you want and be clear about your response.
4 0
3 years ago
Wendy wants to learn how to write a computer program that can get the dimensions of a rectangle from the user and then tell the
otez555 [7]

Answer:

algorithms for finding the area

Explanation:

you need algorithms to find out any computer input information.

4 0
3 years ago
Read 2 more answers
Use semaphore(s) to solve the following problem. There are three processes: P1, P2, and P3. Each process Pi has a segment of cod
Lisa [10]

Answer:

See explaination

Explanation:

Here we will use two semaphore variables to satisfy our goal

We will initialize s1=1 and s2=1 globally and they are accessed by all 3 processes and use up and down operations in following way

Code:-

s1,s2=1

P1 P2 P3

P(s1)

P(s2)

C1

V(s2) .

P(s2). .

. C2

V(s1) .

P(s1)

. . C3

V(s2)

Explanation:-

The P(s1) stands for down operation for semaphore s1 and V(s1) stands for Up operation for semaphore s1.

The Down operation on s1=1 will make it s1=0 and our process will execute ,and down on s1=0 will block the process

The Up operation on s1=0 will unblock the process and on s1=1 will be normal execution of process

Now in the above code:

1)If C1 is executed first then it means down on s1,s2 will make it zero and up on s2 will make it 1, so in that case C3 cannot execute because P3 has down operation on s1 before C3 ,so C2 will execute by performing down on s2 and after that Up on s1 will be done by P2 and then C3 can execute

So our first condition gets satisfied

2)If C1 is not executed earlier means:-

a)If C2 is executed by performing down on S2 then s2=0,so definitely C3 will be executed because down(s2) in case of C1 will block the process P1 and after C3 execute Up operation on s2 ,C1 can execute because P1 gets unblocked .

b)If C3 is executed by performing down on s1 then s1=0 ,so definitely C2 will be executed now ,because down on s1 will block the process P1 and after that P2 will perform up on s1 ,so P1 gets unblocked

So C1 will be executed after C2 and C3 ,hence our 2nd condition satisfied.

4 0
3 years ago
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