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andreev551 [17]
3 years ago
14

I need someone to help. This is the links I found and I'm going through the same thing. I live in small town where they think I'

m crazy. I want this off of me.
https://gajitz.com/forget-bluetooth-human-body-is-efficient-data-transmitter/
Computers and Technology
1 answer:
Hunter-Best [27]3 years ago
8 0

Huh what exactly do you need help with

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“What does it mean when we “rework” copyrighted material?”
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We use ideas and thoughts that have already been protected by the law which is called Copyright. We reworking copyright material means we are taking somebody's ideas and calling it ours. Which is not right by the law.
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Find the double word-length 2's complement representation of each of the following decimal numbers:a. 3874
miss Akunina [59]

Answer:

-3874₁₀ = 1111 1111 1111 1111 1111 1111 1101 1110₂

Explanation:

2's complement is a way for us to represent negative numbers in binary.

To get 2's complement:

1. Invert all the bits

2. Add 1 to the inverted bits

Summary: 2's complement = -N = ~N + 1

1. Inverting the number

3874₁₀ = 1111 0010 0010₂

~3874₁₀ = 0000 1101 1101₂

2. Add 1 to your inverted bits

~3874₁₀ + 1 = 0000 1101 1101₂ + 1

= 0000 1101 1110₂

You can pad the most signigicant bits with 1's if you're planning on using more bits.

so,

12 bits                          16 bits

0000 1101 1110₂  = 1111 0000 1101 1110₂

They asked for double word-length (a fancy term for 32-bits), so pad the left-most side with 1s' until you get a total of 32 bits.

           32 bits

= 1111 1111 1111 1111 1111 1111 1101 1110

7 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

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Which programming scenario would most likely involve this array block?<br><br>SOMEONE PLEASE HELPPP​
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Answer:

answer is D

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as the block returns the number of array members, the most likely scenario is the last one.

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The first one and last one
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