A research team is studying parallel computing. They want to run parallel processes without having to use multiple processors. H
2 answers:
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Answer:
The entire program is:
#include <iostream>
using namespace std;
int main() {
int userInt;
double userDouble;
char userChar;
string userString;
cout<<"Enter integer:"<<endl;
cin>>userInt;
cout<<"Enter double:"<<endl;
cin>>userDouble;
cout<<"Enter character:"<<endl;
cin>>userChar;
cout<<"Enter string:"<<endl;
cin>>userString;
cout<<userInt<<" "<<userDouble<<" "<<userChar<<" "<<userString<<endl;
cout<<endl;
cout<<userInt<<" "<<userDouble<<" "<<userChar<<" "<<userString<<endl<<userString<<" "<<userChar<<" "<<userDouble<<" "<<userInt<<endl;
cout<<endl;
cout<<userInt<<" "<<userDouble<<" "<<userChar<<" "<<userString<<endl<<userString<<" "<<userChar<<" "<<userDouble<<" "<<userInt<<endl<<userDouble<<" cast to an integer is "<<(int)userDouble;
return 0; }
The program in C language:
#include <stdio.h>
int main() {
int userInt;
double userDouble;
char userChar;
char userString[50];
printf("Enter integer: \n");
scanf("%d", &userInt);
printf("Enter double: \n");
scanf("%lf", &userDouble);
printf("Enter character: \n");
scanf(" %c", &userChar);
printf("Enter string: \n");
scanf("%s", userString);
printf("%d %lf %c %s\n", userInt, userDouble, userChar, userString);
printf("\n");
printf("%d %lf %c %s\n%s %c %lf %d \n", userInt, userDouble, userChar, userString, userString, userChar, userDouble, userInt);
printf("\n");
printf("%d %lf %c %s\n%s %c %lf %d\n%lf cast to an integer is %d \n", userInt, userDouble, userChar, userString, userString, userChar, userDouble, userInt, userDouble, (int)userDouble); }
Explanation:
Lets do the program step by step:
1) Prompt the user to input an integer, a double, a character, and a string, storing each into separate variables. Then, output those four values on a single line separated by a space:
Solution:
The program is:
#include <iostream> //to use input output functions
using namespace std; //to identify objects cin cout
int main() { //start of main method
//declare an integer, a double, a character and a string variable
int userInt; //int type variable to store integer
double userDouble; //double type variable to store double precision floating point number
char userChar; //char type variable to store character
string userString; //string type variable to store a string
cout<<"Enter integer:"<<endl; //prompts user to enter an integer
cin>>userInt; //reads the input integer and store it to userInt variable
cout<<"Enter double:"<<endl; //prompts user to enter a double type value
cin>>userDouble; //reads the input double value and store it to userDouble variable
cout<<"Enter character:"<<endl; //prompts user to enter a character
cin>>userChar; //reads the input character and store it to userChar variable
cout<<"Enter string:"<<endl; //prompts user to enter a string
cin>>userString; //reads the input string and store it to userString variable
cout<<userInt<<" "<<userDouble<<" "<<userChar<<" "<<userString<<endl; //output the values on a single line separated by space
So the output of the entire program is:
Enter integer: 99 Enter double: 3.77 Enter character: z Enter string: Howdy 99 3.77 z Howdy
(2) Extend to also output in reverse.
Now the above code remains the same but add this output (cout) statement at the end:
cout<<userString<<" "<<userChar<<" "<<userDouble<<" "<<userInt;
Now the output with the same values given as input is:
Enter integer: 99 Enter double: 3.77 Enter character: z Enter string: Howdy
99 3.77 z Howdy Howdy z 3.77 99
(3) Extend to cast the double to an integer, and output that integer.
The rest of the code remains the same but add the following output (cout) statement in the end:
cout<<userDouble<<" cast to an integer is "<<(int)userDouble;
Now the output with the same values given as input is:
Enter integer: 99 Enter double: 3.77 Enter character: z Enter string: Howdy 99 3.77 z Howdy Howdy z 3.77 99 3.77 cast to an integer is 3
Answer:
<h2>Function 1:</h2>#include <stdio.h> //for using input output functions
// start of the function PrintPopcornTime body having integer variable //bagOunces as parameter
void PrintPopcornTime(int bagOunces){
if (bagOunces < 3){ //if value of bagOunces is less than 3
printf("Too small"); //displays Too small message in output
printf("\n"); } //prints a new line
//the following else if part will execute when the above IF condition evaluates to //false and the value of bagOunces is greater than 10
else if (bagOunces > 10){
printf("Too large"); //displays the message: Too large in output
printf("\n"); //prints a new line }
/*the following else part will execute when the above If and else if conditions evaluate to false and the value of bagOunces is neither less than 3 nor greater than 10 */
else {
/* The following three commented statements can be used to store the value of bagOunces * 6 into result variable and then print statement to print the value of result. The other option is to use one print statement printf("%d",bagOunces * 6) instead */
//int result;
//result = bagOunces * 6;
//printf("%d",result);
printf("%d",bagOunces * 6); /multiplies value of bagOunces to 6
printf(" seconds");
// seconds is followed with the value of bagOunces * 6
printf("\n"); }} //prints a new line
int main(){ //start of main() function body
int userOunces; //declares integer variable userOunces
scanf("%d", &userOunces); //reads input value of userOunces
PrintPopcornTime(userOunces);
//calls PrintPopcornTime function passing the value in userOunces
return 0; }
Explanation:
<h2>Function 2: </h2>#include <stdio.h> //header file to use input output functions
// start of the function PrintShampooInstructions body having integer variable numCycles as parameter
void PrintShampooInstructions(int numCycles){
if(numCycles < 1){
//if conditions checks value of numCycles is less than 1 or not
printf("Too few."); //prints Too few in output if the above condition is true
printf("\n"); } //prints a new line
//else if part is executed when the if condition is false and else if checks //value of numCycles is greater than 4 or not
else if(numCycles > 4){
//prints Too many in output if the above condition is true
printf("Too many.");
printf("\n"); } //prints a new line
//else part is executed when the if and else if conditions are false
else{
//prints "N: Lather and rinse." numCycles times, where N is the cycle //number, followed by Done
for(int N = 1; N <= numCycles; N++){
printf("%d",N);
printf(": Lather and rinse. \n");}
printf("Done.");
printf("\n");} }
int main() //start of the main() function body
{ int userCycles; //declares integer variable userCycles
scanf("%d", &userCycles); //reads the input value into userCycles
PrintShampooInstructions(userCycles);
//calls PrintShampooInstructions function passing the value in userCycles
return 0;}
I will explain the for loop used in PrintShampooInstructions() function. The loop has a variableN which is initialized to 1. The loop checks if the value of N is less than or equal to the value of numCycles. Lets say the value of numCycles = 2. So the condition evaluates to true as N<numCycles which means 1<2. So the program control enters the body of loop. The loop body has following statements. printf("%d",N); prints the value of N followed by
printf(": Lather and rinse. \n"); which is followed by printf("Done.");
So at first iteration:
printf("%d",N); prints 1 as the value of N is 1
printf(": Lather and rinse. \n"); prints : Lather and rinse and prints a new line \n.
As a whole this line is printed on the screen:
1: Lather and rinse.
Then the value of N is incremented by 1. So N becomes 2 i.e. N = 2.
Now at second iteration:
The loop checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to true as N<numCycles which means 2=2. So the program control enters the body of loop.
printf("Done."); prints Done after the above two lines.
printf("%d",N); prints 2 as the value of N is 2
printf(": Lather and rinse. \n"); prints : Lather and rinse and prints a new line \n.
As a whole this line is printed on the screen:
2: Lather and rinse.
Then the value of N is incremented by 1. So N becomes 2 i.e. N = 3.
The loop again checks if the value of N is less than or equal to the value of numCycles. We know that the value of numCycles = 2. So the condition evaluates to false as N<numCycles which means 3>2. So the loop breaks.
Now the next statement is:
printf("Done."); which prints Done on the screen.
So as a whole the following output is displayed on the screen:
1: Lather and rinse.
2: Lather and rinse.
Done.
The programs along with their outputs are attached.
