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3 years ago

The code size of 2-address instruction is________________.? 5 bytes? 7 bytes? 3 bytes? 2 bytes

Computers and Technology

1 answer:

Finger [1]3 years ago

5 0

Answer:

7 bytes

Explanation:

2 Address Instruction

The 2 address instruction consist 3 components in the format.

One is opcode,other two are addresses of destination and source.

Example-

load b,c | Opcode destination address,source address

add a,d | Opcode destination address,source address

sub c,f | Opcode destination address,source address

Opcode consists of 1 bytes whereas destination address and source address consist of 3 bytes each.

(1+3+3) bytes=7 bytes

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3 years ago

Write a program with the total change amount as an integer input, and output the change using the fewest coins, one coin type pe

riadik2000 [5.3K]

Answer:

In Python:

cents = int(input("Cents: "))

dollars = int(cents/100)

quarters = int((cents - 100*dollars)/25)

dimes = int((cents - 100*dollars- 25*quarters)/10)

nickels = int((cents - 100*dollars- 25*quarters-10*dimes)/5)

pennies = cents - 100*dollars- 25*quarters-10*dimes-5*nickels

if not(dollars == 0):

if dollars > 1:

print(str(dollars)+" dollars")

else:

print(str(dollars)+" dollar")

if not(quarters == 0):

if quarters > 1:

print(str(quarters)+" quarters")

else:

print(str(quarters)+" quarter")

if not(dimes == 0):

if dimes > 1:

print(str(dimes)+" dimes")

else:

print(str(dimes)+" dime")

if not(nickels == 0):

if nickels > 1:

print(str(nickels)+" nickels")

else:

print(str(nickels)+" nickel")

if not(pennies == 0):

if pennies > 1:

print(str(pennies)+" pennies")

else:

print(str(pennies)+" penny")

Explanation:

A prompt to input amount in cents

cents = int(input("Cents: "))

Convert cents to dollars

dollars = int(cents/100)

Convert the remaining cents to quarters

quarters = int((cents - 100*dollars)/25)

Convert the remaining cents to dimes

dimes = int((cents - 100*dollars- 25*quarters)/10)

Convert the remaining cents to nickels

nickels = int((cents - 100*dollars- 25*quarters-10*dimes)/5)

Convert the remaining cents to pennies

pennies = cents - 100*dollars- 25*quarters-10*dimes-5*nickels

This checks if dollars is not 0

if not(dollars == 0):

If greater than 1, it prints dollars (plural)

if dollars > 1:

print(str(dollars)+" dollars")

Otherwise, prints dollar (singular)

else:

print(str(dollars)+" dollar")

This checks if quarters is not 0

if not(quarters == 0):

If greater than 1, it prints quarters (plural)

if quarters > 1:

print(str(quarters)+" quarters")

Otherwise, prints quarter (singular)

else:

print(str(quarters)+" quarter")

This checks if dimes is not 0

if not(dimes == 0):

If greater than 1, it prints dimes (plural)

if dimes > 1:

print(str(dimes)+" dimes")

Otherwise, prints dime (singular)

else:

print(str(dimes)+" dime")

This checks if nickels is not 0

if not(nickels == 0):

If greater than 1, it prints nickels (plural)

if nickels > 1:

print(str(nickels)+" nickels")

Otherwise, prints nickel (singular)

else:

print(str(nickels)+" nickel")

This checks if pennies is not 0

if not(pennies == 0):

If greater than 1, it prints pennies (plural)

if pennies > 1:

print(str(pennies)+" pennies")

Otherwise, prints penny (singular)

else:

print(str(pennies)+" penny")

4 0

3 years ago

The speed of sound depends on the material the sound is passing through. Below is the approximate speed of sound (in feet per se

german

Answer:

The Java Program for the given problem is as below. Directly copy the code and run it on your machine.

Explanation:

Refer the Screenshots attached for the output.

import java.io.BufferedReader;

import java.io.InputStreamReader;

public class TheSpeedOfSound {

public static void main(String[] s)

{

String medium;

double distance;

double time;

try{

BufferedReader choice = new BufferedReader(new InputStreamReader(System.in));

System.out.println("Enter one of the following: air, water, or steel: ");

medium = choice.readLine(); // reading input i.e. air, water or steel

//check for air water and steel

if (medium.equalsIgnoreCase("air") || medium.equalsIgnoreCase("water") || medium.equalsIgnoreCase("steel")){

System.out.println("Enter the distance the sound wave will travel: ");

distance = Double.parseDouble(choice.readLine()); // read distance value if it is air water or steel

switch (medium)

{

//if medium is air

case "air":

time = distance/1100;

System.out.print("It will take " + time + " seconds.");

break;

//if medium is water

case "water":

time = distance/4900;

System.out.print("It will take " + time + " seconds.");

break;

//if medium is steel

case "steel":

time = distance/16400;

System.out.print("It will take " + time + " seconds.");

break;

}

else{

System.out.print("Sorry, you must enter air, water, or steel.");

}

catch(Exception e){

e.printStackTrace();

}

8 0

3 years ago