A citizen of any group both

; This program replaces the value in R0 with its absolute value. ; ; It does NOT modify any other registers! ; .ORIG x0200 START

olasank [31]

Using the knowledge in computational language in LC-3 Assembly it is possible to write a code that replaces the value in R0 with its absolute value

<h3>Writting the code </h3>

.ORIG x0200

START

AND R0, R0, #0 ; copy R0 to itself to set the condition codes based on R0;

; i.e performing addition operation with Zero option to set the flags

BRzp DONE ; if R0 is NON-NEGATIVE, skip the negation (already correct);

; Branch to DONE if number is poistive

NOT R0, R0 ; R0 is negative, so negate it i.e taking 2's complement

ADD R0, R0, #1 ; R0 = -R0 is performed successfully

DONE BR START

.END

See more about LC-3 Assembly at brainly.com/question/12978370

#SPJ1

4 0

2 years ago

) how many bits are used for host number on the child network (subnet) , b) how many usable addresses can exist on this child ne

vovikov84 [41]

Answer:

Machine’s IP=126.127.85.170, Machine’s Netmask=/27, Parent’s Netmask=255.255.240.0 .

a) Machine's Netmask = /27 : therefore no. of remaining bits for hosts =32-27 = 5 bits.

b) No. of usable addresses in the child network = 25 -2 = 32-2 =30 [Since first(network ID of the machine) and last ip (broadcast address of the machine ) addresses are not used ]

c) first usable address is on this child network (subnet) =

First, find out the network id of machine can be found out by doing bitwise AND machine's IP and Machine's subnet mask :

01111110. 01111111.01010101.10101010 (IP)

11111111. 11111111 .11111111 .11100000 (Subnet Mask)

01111110. 01111111. 01010101.10100000 (Network ID )

first usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10100001

: 126.127.85.161

d) what the last usable address is on this child network (subnet) : 01111110. 01111111. 01010101.10111110

: 126.127.85.190

e) what the child network’s (subnet's) broadcast address is :

In the directed broadcast address , all the host bits are 1. Therefore, broadcast address :

01111110. 01111111. 01010101.10111111 (126.127.85.191)

f) what the child network's (subnet's) network number is : Network ID has already been calcuulated in part c .

01111110. 01111111. 01010101.10100000 (126.127.85.160)

6 0

3 years ago

Write the addReview method, which adds a single product review, represented by a ProductReview object, to the ReviewCollector ob

maw [93]

Answer:

Explanation:

namespace Jeroen\ReviewIntegration\Observer;

use Magento\Framework\Event\ObserverInterface;

class ProductReview implements ObserverInterface

{

protected $_storeManager;

protected $_request;

public function __construct(

\Magento\Store\Model\StoreManagerInterface $storeManager,

\Magento\Framework\App\Request\Http $request

) {

$this->_storeManager = $storeManager;

$this->_request = $request;

}

public function execute(\Magento\Framework\Event\Observer $observer)

{

return 'test';

}

5 0

3 years ago

Programming in https<br>

kakasveta [241]

What?? explain your question

3 0

2 years ago

Consider two different implementations, M1 and M2, of the same instruction set. There are three classes of instructions (A, B, a

ANTONII [103]

Answer:

a) the average CPI for machine M1 = 1.6

the average CPI for machine M2 = 2.5

b) M1 implementation is faster.

c) the clock cycles required for both processors.52.6*10^6.

Explanation:

(a)

The average CPI for M1 = 0.6 x 1 + 0.3 x 2 + 0.1 x 4

= 1.6

The average CPI for M2 = 0.6 x 2 + 0.3 x 3 + 0.1 x 4

= 2.5

(b)

The average MIPS rate is calculated as: Clock Rate/ averageCPI x 10^6

Given 80MHz = 80 * 10^6

The average MIPS ratings for M1 = 80 x 10^6 / 1.6 x 10^6

= 50

Given 100MHz = 100 * 10^6

The average MIPS ratings for M2 = 100 x 10^6 / 2.5 x 10^6

= 40

c)

Machine M2 has a smaller MIPS rating

Changing instruction set A from 2 to 1

The CPI will be increased to 1.9 (1*.6+3*.3+4*.1)

and hence MIPS Rating will now be (100/1.9)*10^6 = 52.6*10^6.

7 0

3 years ago