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AfilCa [17]
3 years ago
7

Consider the function definition void Demo( int& intVal, float floatVal ) { intVal = intVal * 2; floatVal = float(intVal) +

3.5; } Suppose that the caller has variables myInt and myFloat whose values are 20 and 4.8, respectively. What are the values of myInt and myFloat after return from the following function call? Demo(myInt, myFloat);
Computers and Technology
1 answer:
Fynjy0 [20]3 years ago
5 0

Answer:

myInt=40

myFloat=4.8

Explanation:

First look at the function definition .It has two arguments intVal is passed by reference while floatVal is passed by value.So the changes done on the myInt variable will be reflected on the original argument because when a variable is passed by reference the the changes are reflected on the original argument but when a variable is passed by value the function created a duplicate copy of it and work on them so changes are not reflected on the original argument.So myInt will get doubled while myFloat will remain the same.

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The development team recently moved a new application into production for the accounting department. After this occurred, the Ch
Gnesinka [82]

Answer:

B

Explanation:

User Acceptance Testing (UAT) is the testing done by end users who is going to use that application. In these testing they will test  whether the application meets all the functional requirements of the end user. If it passed then only the application will move to the production. This will ensure that the application meets the user requirements before it put into the usage.

The Answer is B

5 0
3 years ago
You are building your own computer, and you are ready to select
KiRa [710]

Answer:

D

Explanation:

It's because you need a system software so the user can understand what the hardware is saying because the hardware only uses binary codes.

7 0
3 years ago
Develop a plan to design and finally implement a set of functions using C++ that would implement the IEEE standard. Phase 1 will
AnnZ [28]

Answer:

See explaination code

Explanation:

#include<iostream>

#include<math.h>

using namespace std;

typedef union {

float number;

struct

{

// Order is important.

// Here the members of the union data structure

// use the same memory (32 bits).

// The ordering is taken

// from the LSB to the MSB.

unsigned int mantissa : 23;

unsigned int exponent : 8;

unsigned int sign : 1;

} Raw;

} MyFloat;

void printBinary(int n, int i)

{

// Prints the binary representation

// of a number n up to i-bits.

int k;

for (k = i - 1; k >= 0; k--) {

if ((n >> k) & 1)

cout << "1";

else

cout << "0";

}

}

void decToHex(int n){

// char array to store hexadecimal number

char hexaDeciNum[100];

// counter for hexadecimal number array

int i = 0;

while(n!=0)

{

// temporary variable to store remainder

int temp = 0;

// storing remainder in temp variable.

temp = n % 16;

// check if temp < 10

if(temp < 10)

{

hexaDeciNum[i] = temp + 48;

i++;

}

else

{

hexaDeciNum[i] = temp + 55;

i++;

}

n = n/16;

}

// printing hexadecimal number array in reverse order

for(int j=i-1; j>=0; j--)

cout << hexaDeciNum[j];

}

void floatBinary(float f){

long double binaryTotal, binaryFrac = 0.0, frac, fracFractor = 0.1;

long int integer, binaryInt = 0;

long int p = 0, rem, temp;

//separate the integer part from the input floating number

integer = (int)f;

//separate the fractional part from the input floating number

frac = f - integer;

//loop to convert integer part to binary

while (integer != 0) {

rem = integer % 2;

binaryInt = binaryInt + rem *pow(10, p);

integer = integer / 2;

p++;

}

//loop to convert fractional part to binary

while (frac != 0) {

frac = frac * 2;

temp = frac;

binaryFrac = binaryFrac + fracFractor * temp;

if (temp == 1)

frac = frac - temp;

fracFractor = fracFractor / 10;

}

cout << binaryInt + binaryFrac;

}

int findDecimal(float number){

int nfloor = number;

float nfloat = number - nfloor;

int nfloatfloor;

do {

nfloat *= 10;

nfloatfloor = nfloat;

} while (nfloat > nfloatfloor);

return nfloatfloor;

}

void first(float number){

if(number < 0)

cout << "SIGN BIT IS (1) SINCE NUMBER IS NEGATIVE" << endl;

else

cout << "SIGN BIT IS (0) SINCE NUMBER IS POSITIVE" << endl;

}

void second(float number){

cout << "INTEGER PART IN BASE-10:" << int(number) << " AND IN BINARY:";

printBinary(int(number),16);

cout << endl;

}

void third(float number){

cout << "DECIMAL PART IN BASE-10:" << findDecimal(number) << " AND IN BINARY:";

printBinary(findDecimal(number),16);

cout << endl;

}

void fourth(float number){

cout << "ENTERED NUMBER IN BASE-10:" << number << " AND IN BINARY:";

floatBinary(number);

cout << endl;

}

void fifth(MyFloat myfloat){

cout << "MANTISA IN BINARY:";

printBinary(myfloat.Raw.mantissa,32);

}

void sixth(MyFloat myfloat){

cout << "EXPONENT IN BASE-10:" << myfloat.Raw.exponent << " AND IN BINARY:";

printBinary(myfloat.Raw.exponent,8);

cout << endl;

}

void seventh(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

printBinary(myfloat.Raw.exponent,8);

cout << " | ";

printBinary(myfloat.Raw.mantissa,32);

cout << endl;

}

void eigth(MyFloat myfloat){

cout << myfloat.Raw.sign << " | ";

decToHex(myfloat.Raw.exponent);

cout << " | ";

decToHex(myfloat.Raw.mantissa);

cout << endl;

}

int main(){

float number;

cout << "PLEASE ENTER A NUMBER TO DISPLAY THE IEEE 754 FLOATING POINT OPTIONS" << endl;

cin >> number;

MyFloat myfloat;

myfloat.number = number;

cout << "PLEASE CHOOSE ONE OF THE FOLLOWING OPERATIONS" << endl;

cout << " 1. DISPLAY THE SIGN BIT VALUE" << endl;

cout << " 2. DISPLAY THE INTEER PART IN BOTH BASE-10 AND CINARY FORMATS" << endl;

cout << " 3. DISPLAY THE DECIMAL PART IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 4. DISPLAY THE NUMBER ENTERED IN BOTH BASE-10 AND BINARY FORMATS" << endl;

cout << " 5. DISPLAY THE MANTISA IN BINARY FORMATS" << endl;

cout << " 6. DISPLAY THE EXPONENT IN BORH BASEE-10 AND BINARY FORMATS" << endl;

cout << " 7. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

cout << " 8. DISPLAY THE IEEE 754 SINGLE PRECISION BINARY LAYOUT" << endl;

int choice;

cin >> choice;

switch(choice){

case 1:first(number);

break;

case 2:second(number);

break;

case 3:third(number);

break;

case 4:fourth(number);

break;

case 5:fifth(myfloat);

break;

case 6:sixth(myfloat);

break;

case 7:seventh(myfloat);

break;

case 8:eigth(myfloat);

break;

default:cout << "ENTER VALID CHOICE" << endl;

}

}

Refer to attachment please for onscreen look.

6 0
3 years ago
Create a Boolean function odd_number_digits(n) that returns True when a positive integer has an odd number of digits. (You may a
Snowcat [4.5K]

Answer:

Following are the code to this question:

import java.util.*;//import package for user input

public class Main//defining class main

{

   public static boolean odd_number_digits(int n)//defining boolean method odd_number_digits

   {

       if(n>0 && n%2!=0)//defining if block that check value is positive and odd number

       {

       return true;//return value true

       }

       else//defining else block

       {

           return false;//return false value

       }

   }

   public static void  sum_odd_digits(int n)//defining a method sum_odd_digits

   {

       int sum=0,i;//defining integer variable

       for(i=0;i<=n;i++)//defining for loop

       {

           if(i%2!=0)//defining if block for odd number

           {

               sum=sum+i;//add odd number

           }

       }

      System.out.print(sum);//use print method to print sum value

   }

public static void main(String[] args) //defining main method

   {

       Scanner ox=new Scanner(System.in);//creating Scanner object

       int n= ox.nextInt();//defining integer variable for input value

       System.out.print(odd_number_digits(n)+ "\n");//use print method to call method

       System.out.println("Sum of odd numbers: ");//print message

       sum_odd_digits(n);//calling method

   }

}

Output:

please find the attachment.

Explanation:

In the above code, two methods "odd_number_digits and sum_odd_digits" are defined in which the first method return type is boolean because it will true or false value, and the second method returns the sum of odd numbers.

In the "odd_number_digits" method,  an integer variable passes as an argument and inside the method, if block is used that check value is a positive and odd number then it will return a true value.

In the "sum_odd_digits" method, it accepts an integer parameter "n", and define integer variable "sum" inside the method, which uses the for loop, inside the loop if block is used that counts odd numbers and adds its in sum and print its value.

8 0
3 years ago
Mac os is based on the ____ operating system.
ankoles [38]
Mac os is based on the UNIX operating system.
4 0
3 years ago
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