Concept explainers
Consider the following code (and assume that it is embedded in a complete and correct
char next;
int count = 0;
cout <<“Enter a line of input:\n”;
cin. get(next);
while (next != ‘\n’)
{
if ((count % 2) == 0) True if count is even
cout << next;
count ++;
cin.get(next) ;
}
If the dialogue begins as follows, what will be the next line of output?
Enter a line of input:
abcdef gh
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- (Numerical) Write a program that tests the effectiveness of the rand() library function. Start by initializing 10 counters to 0, and then generate a large number of pseudorandom integers between 0 and 9. Each time a 0 occurs, increment the variable you have designated as the zero counter; when a 1 occurs, increment the counter variable that’s keeping count of the 1s that occur; and so on. Finally, display the number of 0s, 1s, 2s, and so on that occurred and the percentage of the time they occurred.arrow_forward#include <stdio.h>int main(){int d;int s[20],i, r, p, lg=0,m;char c;printf("Enter number of salesman(max 20): ");scanf("%d", &d);for(i=0; i<d; i++){ printf("\n salesman %d sales: ");scanf("%d", &s[i]);}for(i=0; i<d; i++){for(r=i+1; j<d; r++){if(s[i] > s[r]){p= s[i];s[i] = s[r];s[j] = p;}}}printf("\nsalesman lowest to highest: ");for(i=0; i<d; i++){printf("%d\t", s[i]);}for(i=0;i<m;i++){printf("\n highest sales: %d ",lg);if(lg<=s[i])lg=s[i];break;}getch();} >>> the upper part output should be like this enter number of salesman (max 20): 5 salesman 1 500 salesman 2 300 salesman 3 1000 salesman 4 200 salesman 5 1000 in the lower part the lowest to highest the output should become like this salesman 4 200 salesman 2 300 salesman 1 500 salesman 3 1000 salesman 5 1000 highest total sales : 2000arrow_forwardCFG: Example 1 • Draw the CFG for the following code: int f(int n){ } int m = n* n; if (n < 0) else return 0; return m;arrow_forward
- 2- The factorial n! of a positive integer n is defined as n! = 1*2*3 . .. * (n-1) * n Where 0! = 1 Write a function to calculate the factorial of a number. Argument: A number n of type unsigned int. Returns: The factorial n! of type long double. Write two versions of the function, where the factorial is • calculated using a loop calculated recursively Test both functions by outputting the factorials of the numbers 0 to 20.arrow_forwardEnglish (e what is the output of the following program? #include int test (int &, int); main () { int a=10, b=15; test (b, a); test (a,b); cout<arrow_forwardFind an incorrect line in the following code segment: void swap(int *x, int *y) { int t = *x; *x = *y; *y = t;}....int a = 2, b = 3;swap(*a, *b);arrow_forwardGiven x 0, у 3 1, count 3. After the following code is executed, count ++; y=x; if (x < 0) count--; else count++; the value of x isarrow_forward2. Test Scores File: test_scores.py Write pseudocode for the main() part of a program that asks the user to enter 4 test scores between 0 and 100, then displays a JCU grade for each score and also the average test score. When you have written the pseudocode for main, implement your solution in Python code and test it with a range of meaningful data. Remember that we've done the JCU grades question before, so copy your function from that practical code file. Sample Output Score: 3 Score: 50.5 Score: 66 Score: 100 Score 3.0, which is N Score 50.5, which is P Score 66.0, which is C Score 100.0, which is HD The average score was 54.875 Enhancements When you have that working... We asked for 4 scores. Have a look at your code... did you use 4 as a numeric literal or a constant?Change 4 to 3... Did you have to change the program in more than one place?If so, then you've missed one of the things we've taught...As a strong guideline: if you need to use the same literal more than once, you…arrow_forward1. a. Are Count and sum the same? yes or no, why? { int sum = 0; for ( int count = 0; count < 10; count++ ) { System.out.print( count + " " ); sum = sum+count; } } b. { int sum = 0; for ( int count = 0; count < 10; count++ ) { System.out.print( count + " " ); sum = sum+count; } System.out.println("\nAfter the loop count is: " + count ); } c. What is printed, why? class FindIt { private int sum; public FindIt( int sum ) { this.sum = sum; } public void increment( int inc ) { sum = sum + inc; System.out.println("FindIt sum: " + sum ); } } public class MainClass { public static void main ( String[] args) { int sum = 99; FindIt findObj = new FindIt( 34 ); findObj.increment( 6 ); System.out.println("sum: " + sum ); } }arrow_forwardQuestion 1: Please develop a program. The program shows a O(n^2) time complexity. You may design nested loops. When you input a number, the output should be how many times of the loops. For example, if your input is 10, the number of loops should be 100 times, 200 times, 300 times, etc. The number of loops do not have to be an accurate number. Question 2: Please develop a program. The program shows a O(nlgn) time complexity. You may design nested loops. When you input a number, the output should be how many times of the loops. For example, if your input is 10, the number of loops should be 30 times, 60 times, etc. The number of loops do not have to be an accurate number.arrow_forward#include<stdio.h> #include<stdlib.h> main() { int m, sum = 0, counter = 0; int first = 2147483647, second = 2147483647, third = 2147483647, min = 2147483647; double average;printf("Enter an int or -1 to stop:\n"); while (1) {scanf_s("\n%d", &m);if (m == -1) { break; } sum = sum + m; counter++; if (m < first first == second first == third ) { third = second; second = first; first = m; } else if (m < second && m !=first ) { third = second; second = m; } else if (m < third && m != second) { third = m; } else if (first == second == third) { printf("min is: %d", first); }}printf("Sum of value: %d \n", sum); average =(double) sum /(double)counter; printf("Avarege is: %.2lf \n", average); printf("First min is: %d\n", first); printf("Second min is: %d\n", second); printf("Third min is: %d\n", third);system("pause"); } can you run this code pleasearrow_forward#include<stdio.h> #include<conio.h> int mutex=1,full=0,empty=1,x=0; main() { int n; void producer(); void consumer(); int wait(int); int(signal); printf("\n1. Producer \n2. Consumer \n,3. Exit \n"); while(1) { printf("\n Enter the choice:"); scanf("%d", &n); switch(n); { case-1: if((mutex==1)&&(empty!=0)) producer(); else printf("Nuffer is full"); break; case-2: if((mutex==1)&&(full!=0)) consumer(); else print("Buffer is empty"); break; case-3: exit(0); break; } } } int wait(int s) {return(--s); } int signal(int s) { return(++s); } void producer() { mutex = wait(mutex); full = signal(full); empty=wait(empty); x++; printf("\n Producer produces the item %d",x); mutex= signal(mutex); } void consume() { mutex = wait(mutex); full = wait(full); empty = signal(empty); printf("\n Consumer consumes item %d",x); x--; mutex = signal(mutex); } CONVERT THIS C LANGUAGE CODE IN BASH SCRIPT THANKYOUarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- C++ for Engineers and ScientistsComputer ScienceISBN:9781133187844Author:Bronson, Gary J.Publisher:Course Technology Ptr