I'm trying to swap two array elements that are initialized strings (see code below), I thought i knew how to do it but the way I tried doesn't seem to be working. We're not allowed to use the function "swap" that I've seen on many other forum sights with a similar question as mine. Therefore I've used a temporary index variable to swap them but that doesn't seem to be working. Not sure how to fix it and make it work, so my question is how do I do it.
I'm pretty new to programming so the answer may not be as evident to me yet. I've been staring at this for a while and still can't see it, I also tried asking on reddit but they didn't give me a very concise or helpful answer. If you could help me out as to why it wont swap the elements that would be great and if you see any other bugs or improvements I could make please let me know, your feedback is greatly appreciated, thank you!
Code:
#include <iostream>
using namespace std;
void printArray(string names[]) {
for (int i = 0; i < 7; i++) {
cout << names[i] << " ";
}
}
int main() {
int x = 0, a, b;
string answer, name1, name2;
string index;
string names[7] = {"John", "Dave", "Jim", "Amanda", "Kelsey", "Don", "Jennifer"};
printArray(names);
while (x < 1) {
cout << endl << "Do you want to swap students? " << endl;
cin >> answer;
if (answer == "Yes" || answer == "yes") {
cout << "Who would you like to swap?" << endl;
cin >> name1;
for(a = 0; a < 7; a++) {
if (names[a] == name1) {
cout << "Yes, " << name1 << " is in the line." << endl;
}
}
cout << "Who would you like to swap " << name1 << " for?" << endl;
cin >> name2;
for(b = 0; b < 7; b++) {
if (names[b] == name2) {
cout << "Yes, " << name2 << " is in the line!!" << endl;
index = names[a];
names[a] = names[b];
names[b] = index;
printArray(names);
}
}
} else {
cout << endl << "Thanks, please behave now, students!" << endl;
x++;
}
}
return 0;
}
More context:
Print out the current class line-up (i.e., the current contents of the array, in order).
Present the user with a prompt asking if they would like to swap 2 students.
If they say yes, proceed to step 3.
Ask the user for the names of the two students to be swapped.
If the two students are both in the array, swap their positions in the array. If either student is not in the class, print an error message on the console, e.g. "Sorry, you have to pick 2 students who are in the line!".
No matter the outcome of step 4, return to step 1.
While searching for name1 ,the loop runs till 7 no matter if its found or not which messes with your swapping task:
index = names[a]; //a=7
names[a] = names[b]; //a=7
names[b] = index;.
Use break:
if (names[a] == name1) {
cout << "Yes, " << name1 << " is in the line." << endl;
break;
}
The value of a is always 7 while swapping is being done! So it is not working.
Related
I've started to learn how to code in C++ on my spare time, using different sites and apps that someone who has also learned C++ online provided me with. By now, I know the most basic commands. I've tried an exercise given by a program, and I'm given the information that someone is going on a vacation, and needs to know how much baggage he can bring with him. The limit to how many baggages he can carry is 45, and I have to display a different output if the baggages are below, above or the same as the limit (45 baggages). I have done some coding, and I ended up with this:
#include <iostream>
using namespace std;
int main()
{
const int limit = 45;
int bag;
cout << "Please type your number here: ";
cin >> bag;
string yn;
int keep = 0;
if (limit < bag)
{
cout << "You passed the limit." << endl;
};
if (limit == bag)
{
cout << "Just enough." << endl;
};
if (limit > bag)
{
cout << "You got space." << endl;
};
++keep;
while(keep > 0)
{
int keep = 0;
cout << "Do you want to try another number?" << endl;
cin >> yn;
cout << endl;
if(yn == "yes")
{
int bag = 0;
cout << "Please type your number here: ";
cin >> bag;
if (limit < bag)
{
cout << "You passed the limit." << endl;
};
if (limit == bag)
{
cout << "Just enough." << endl;
};
if (limit > bag)
{
cout << "You got space." << endl;
};
}
else
{
return 0;
}
}
}
I have developed it more than needed -as you can see-, out of my own interest in the problem. I have copied and pasted the 3 IF commands as seen above, and I believe that there is an easier way, with less code, do solve this. What I have thought of is if I could go back and execute some line of code again, either from a line and below (e.g. from line 45 and below), or specific lines of code (e.g. from line 45 to line 60).
I would appreciate it if you thought of another way to solve this problem and posted your code below.
Thank you for your reply.
We all started writing our first C++ program at some time, so allow me to give you some additional feedback:
First of all, avoid writing using namespace std;
Secondly, naming - what is bag, limit, keep and yn? Wouldn't it be much easier to read and understand if they were called bagSize, maximumPermittedBagSize, inputFromUser (you don't really need the variable keep, see below)?
Finally, here is a (roughly) refactored version your program, with duplication removed and comments added.
#include <iostream>
int main()
{
const int maximumPermittedBagSize = 45;
// Loops forever, the user exits by typing anything except 'yes' laster
while(true)
{
std::cout << "Please type your number here: " << std::endl;
//Declare (and initialize!) variables just before you need them
int bagSize = 0;
std::cin >> bagSize;
if (bagSize > maximumPermittedBagSize)
{
std::cout << "You passed the limit." << std::endl;
}
else if (bagSize == maximumPermittedBagSize )
{
std::cout << "Just enough." << std::endl;
}
else
{
std::cout << "You got space." << std::endl;
}
std::cout << "Do you want to try another number?" << std::endl;
std::string inputFromUser = "";
std::cin >> inputFromUser;
std::cout << std::endl;
//Leave the loop if the user does not answer yes
if(inputFromUser != "yes")
{
return 0;
}
}
}
You can simply run a while loop and do like this:
#include <iostream>
using namespace std;
int main()
{
const int limit = 45;
int bag;
string yn = "yes";
while(yn == "yes")
{
cout << "Please type your number here: ";
cin >> bag;
if (limit < bag)
{
cout << "You passed the limit." << endl;
}
else if (limit == bag)
{
cout << "Just enough." << endl;
}
else if (limit > bag)
{
cout << "You got space." << endl;
}
cout << "Do you want to try another number?" << endl;
cin >> yn;
cout << endl;
}
}
I'm trying to work through the beginner exercises from a website.
"Requires:
variables, data types, and numerical operators
basic input/output
logic (if statements, switch statements)
loops (for, while, do-while)
arrays
Write a program that asks the user to enter the number of pancakes eaten for breakfast by 10 different people (Person 1, Person 2, ..., Person 10)
Once the data has been entered the program must analyze the data and output which person ate the most pancakes for breakfast."
I'm unsure on how to get the program to call out the person which enters the most number of pancakes eaten? Surely this would need to be done with a key and value, but the requirements state 'arrays' but not 'maps'?
Below is the code I have come up with, but this only outputs the maximum number of pancakes eaten, so not really answering the question!
Thanks so much for any help!
* I've only used 5 people to quicken the process before I know exactly how to do it *
#include <iostream>
using namespace std;
int main()
{
cout << "how many pancakes did you eat for breakfast?" << endl;
int person1, person2, person3, person4, person5;
cout << "Person 1: ";
cin >> person1;
cout << "Person 2: ";
cin >> person2;
cout << "Person 3: ";
cin >> person3;
cout << "Person 4: ";
cin >> person4;
cout << "Person 5: ";
cin >> person5;
int array[5] = {person1, person2, person3, person4, person5};
int temp = 0;
for (int i = 0; i<5; i++)
{
if (array[i] > temp)
{
temp = array[i];
}
}
cout << "The most pancakes eaten was " << temp << "by " << endl;
}
Surely this would need to be done with a key and value
This is not the only way of doing it. Another way is to use an indexed collection with no key, and make an assumption that position k corresponds to a key k that can be computed from a position alone. For example, if you have an array of ten items corresponding to ten people numbered 1 through 10, then the data for a person number k could be stored in the array at position k-1. No keys are required in this situation.
This long explanation means that if you store the best i in addition to best tmp, you'll have your answer after the loop:
int temp = 0;
int res = -1;
for (int i = 0; i<5; i++) {
if (array[i] > temp) {
temp = array[i];
res = i;
}
}
cout << "The most pancakes eaten was " << temp << "by " << (res+1) << endl;
Note the res+1 is printed, not res. This is because arrays are zero-based, while counting is one-based.
This could be further shortened using a common idiom of using the initial element as the current best, and starting your iterations from 1:
int res = 0;
for (int i = 1 ; i<5 ; i++) {
if (array[i] > array[res]) {
res = i;
}
}
cout << "The most pancakes eaten was " << array[res] << "by " << (res+1) << endl;
What if you kept track of the maximum amount of pancakes eaten as you took input?
#include <iostream>
using namespace std;
// To execute C++, please define "int main()"
int main() {
int numPeople = 5;
int maxPancakes = -1;
int maxPerson = -1;
int currentPancakes = -1;
for (int i = 1; i < numPeople; i++) {
cout << "Person " << i << ": ";
cin >> currentPancakes;
if (currentPancakes > max) {
max = currentPancakes;
maxPerson = i;
}
}
cout << "Person " << maxPerson << " ate the most pancakes: " << maxPancakes;
return 0;
}
Note: my c++ is pretty rusty, I haven't tested this solution. Just an idea ;)
Using Map for this question will be an overkill. Array is more than enough. You don't even need to iterate through the array to check who ate the most. The operation for getting the max is actually O(0) because we can update who ate the most as you are entering the values.
int main(){
const int NUM_PEOPLE = 10;
int cakesEaten[10] = {0};
int maxEaten = 0;
int personId = 0;
cout << "How many pancakes eaten by:" << endl;
for(int x=0; x<NUM_PEOPLE; x++){
cout << "person " << (x+1) << ":";
cin >> cakesEaten[x];
if (cakesEaten[x] > maxEaten){
maxEaten = cakesEaten[x];
personId = x;
}
}
cout << "The most pancakes was eaten by person " << personID << endl;
}
You don't need any storage at all.
As the numbers are entered, compare them and store who has the current max, and its value
Starting with fake values is not needed if you use the first person's value as the start value, this way negative values, could be included if entered. That may be nonsensical here, but in general, its a better practice.
Also note if we want people to start at 1, then it makes more sense to start it at 1, then start at 0 and try to remember to always add 1.
This is also very easy to expand to more people, just change total_people
int main() {
const int total_people=5;
cout << "how many pancakes did you eat for breakfast?" << endl;
int what;
cout << "Person 1: ";
cin >> what;
int who=1;
int max_value=what;
for (int person = 2; person <= total_people; ++person) {
cout << "Person " << person << ": ";
cin >> what;
if (what > max_value) {
max_value=what;
who=i;
}
}
cout << "The most pancakes eaten was " << max_value << "by " << who << endl;
}
Heres a function of my program, which keeps crashing. The purpose of the function is the search through the vectors and display what they rank and the frequency, which is read into thier respective vectors. Im guessing its my while loop but im not sure.
The vectors are correctly read in.
void searchNames(vector<string> girlsname,vector<int> girlsfeq,vector<string> boysname,vector<int> boysfeq){
string name1;
string name;
int SIZE = girlsname.size();
int SIZE2 = boysname.size();
while(name1 != "quit"){
int i=0;
int frank=find(girlsname.begin(), girlsname.end(), name1) - girlsname.begin();
int mrank=find(boysname.begin(), boysname.end(), name1) - boysname.begin();
name1.clear();
name.clear();
cout << "Please enter a name: ";
cin >> name1;
cout << endl;
while(i< SIZE && name1 !=girlsname[i++]){
frank++;
}
name.clear();
i=0;
while(i< SIZE2 && name1 !=boysname[i++]){
mrank++;
}
cout << left << setw(15) << "Name" << right << setw(10) << "Girl Rank" << right << setw(10) << "Frequency" << right << setw(10) << "Boy Rank"<< right << setw(10) << "Frequency" << endl;
if(frank != SIZE && mrank != SIZE2){
cout << left << setw(15) << name1 << right << setw(10) << frank << right << setw(10) << girlsfeq[frank] << right << setw(10) << mrank << right << setw(10) << boysfeq[mrank] << endl;
}
}
First, in any of your loops, you don't change the value of i, so they are potentially infinite.
Second, if I understand what you're trying to do, you have paired vectors and you search for an index on which the name is in the first vector and then use it to get the value from second vector. You can use std::find function for that:
int frank = find(girlsname.begin(), girlsname.end(), name1) - girlsname.begin();
Now, you can access the girlsfreq[frank] value. If the name is not present in the vector, the frank equals size of the vector, which you have to check for. Don't forget to add include:
#include <algorithm>
I am not able to reproduce your crash. Instead, I am able to reproduce an infinite loop. Taking a closer look to your while loops, it becomes evident: the value of i is never changing. If it cannot find a name match, it will keep incrementing the value of frank. So, a fix to this problem would be:
while(i < SIZE && name1 != name) {
name = girlsname[i];
frank++;
i++;
}
But it can be improved: you don't need that assignment. Also, you can use the advantage of the postfix increment:
while(i < SIZE && name1 != girlsname[i++]){
frank++;
}
Doing this, you also remove the necessity of name and name1.
as recommended I've been working through the book 'Jumping into c++'. I'm currently on problem 7 of chapter 5 and although I have produced the code that appears to do what is asked of me I was hoping someone might be able to take a look and tell me if I've implemented any 'bad' practice (Ideally I don't want to be picking up bad habits already).
Secondly, it also says 'try making a bar graph that shows the results properly scaled to fit on your screen no matter how many results were entered'. Again, the code below produces a horizontal bar graph but I'm not convinced that if I had say 10000 entries (I guess I could verify this by adding an additional for loop) that it would scale according. How would one go about applying this? (such that it always properly scales regardless of how many entries).
I should probably point out at this point that I have not covered topics such as arrays, pointers and classes as of yet in case anyone was curious as to why I didn't just create a class called 'vote' or something.
One final thing... I don't have a 'return 0' in my code, is this a problem? I find it slightly confusing as to what exactly the point of having return 0 is. I know that it's to do with making sure your code is running properly but it seems sort of redundant?
Thanks in advance!
#include <iostream>
using namespace std;
int main()
{
int option;
int option_1 = 0;
int option_2 = 0;
int option_3 = 0;
cout << "Which is your favourite sport?" << endl;
cout << "Tennis.. 1" << endl;
cout << "Football.. 2" << endl;
cout << "Cricket.. 3" << endl;
cin >> option;
while(option != 0)
{
if(option == 1)
{
option_1++;
}
else if(option ==2)
{
option_2++;
}
else if(option ==3)
{
option_3++;
}
else if(option > 3 || option < 0)
{
cout << "Not a valid entry, please enter again" << endl;
}
else if(option ==0)
{
break;
}
cout << "Which is your favourite sport?" << endl;
cout << "Tennis.. 1" << endl;
cout << "Football.. 2" << endl;
cout << "Cricket.. 3" << endl;
cin >> option;
}
cout << "Option 1 (" << option_1 << "): ";
for(int i = 0; i < option_1; i++)
{
cout << "*";
}
cout << "" << endl;
cout << "Option 2 (" << option_2 << "): ";
for(int i = 0; i < option_2; i++)
{
cout << "*";
}
cout << "" << endl;
cout << "Option 3 (" << option_3 << "): ";
for(int i = 0; i < option_3; i++)
{
cout << "*";
}
}
About the return 0 in main : it's optional in C++.
About your code:
You have a ton of if / else if blocks, you should replace them with a switch. A switch statement is more compact, readable, and may be a little bit faster at runtime. It's not important at this point, but it's pretty good practice to know where to put a switch and where to use regular if.
You have one big function, it's really bad. You should break your code into small, reusable pieces. That's something called DRY (Don't repeat Yourself): if you are copy-pasting code, you're doing something wrong. For example, your sport list appears 2 times in your code, you should move it in a separate function.
You wrote cout << "" << endl;, I think you don't really understand how std::cout work. std::cout is an object representing the standard output of your program. You can use operator<< to pass values to this standard output. std::endl is one of these values you can pass, strings are, too. So you can just write cout << endl;, no need for an empty string.
Please learn how to use arrays, either raw ones or std::array. This is a pretty good example of a program which can be refactored using arrays.
Here is a more readable, cleaner version of your code:
#include <iostream>
int prompt_option()
{
int option;
while (true)
{
std::cout << "Which is your favourite sport?" << std::endl;
std::cout << "Tennis.. 1" << std::endl;
std::cout << "Football.. 2" << std::endl;
std::cout << "Cricket.. 3" << std::endl;
std::cin >> option;
if (option >= 0 && option <= 3)
return option;
else
std::cout << "Not a valid entry, please enter again" << std::endl;
}
}
void display_option(int number, int value)
{
std::cout << "Option " << number << " (" << value << "): ";
while (value--)
std::cout << '*';
std::cout << std::endl;
}
int main()
{
int option;
int values[3] = {0};
while (true)
{
option = prompt_option();
if (option)
values[option - 1]++;
else
break;
}
for (int i = 0; i < 3; i++)
display_option(i + 1, values[i]);
}
You have too much if else, it messy.
check out the code bellow, muc shorter, cleaner and efficent.
I hope it helps.
#include <iostream>
using namespace std;
int main()
{
int choice;
int soccer=0, NFL=0 ,formula1=0;
while(choice != 0){
cout<<"Please choose one of the following for the poll"<<endl;
cout<<"press 1 for soccer, press 2 for NFL, press 3 for formula 1"<<endl;
cout<<"Press 0 to exit"<<endl;
cin>>choice;
if(choice==1){
soccer++;
}
else if(choice==2){
NFL++;
}
else if(choice == 3){
formula1++;
}
else{
cout<<"Invalid entry, try again"<<endl;
}
cout<<"soccer chosen "<<soccer<<" times.";
for(int i=0; i<soccer; i++){
cout<<"*";
}
cout<<endl;
cout<<"NFL chosen "<<NFL<<" times.";
for(int j=0; j<NFL; j++){
cout<<"*";
}
cout<<endl;
cout<<"formula1 chosen "<<formula1<<" times.";
for(int c=0; c<formula1; c++){
cout<<"*";
}
cout<<endl;
}
return 0;
}
I'm new to this site and programming in C++ language this semester.
I have been really trying for 2 days and have asked classmates but they do not know either. A classmate said to use the 2D arrays but I don't know what that is and my professor has not gone over 2D arrays.
I am stuck and would really appreciate help.
The input file contains this:
Plain_Egg 1.45
Bacon_and_Egg 2.45
Muffin 0.99
French_Toast 1.99
Fruit_Basket 2.49
Cereal 0.69
Coffee 0.50
Tea 0.75
idk how to display all the "users" orders
Basically a receipt, like he orders this and how many, then ask "u want anything else?", then take the order number and how many again, THEN at the end give back a receipt that looks like this
bacon_and_eggs $2.45
Muffin $0.99
Coffee $0.50
Tax $0.20
Amount Due $4.14
Here is my code:
// Libraries
#include <iostream>
#include <string>
#include <iomanip>
#include <fstream>
#include <cmath>
using namespace std;
//structures
struct menuItemType {
string itemName;
double itemCost;
};
// Prototypes
void header();
void readData(menuItemType menu[]);
void Display(menuItemType menu[]);
int main() {
header();
menuItemType menu [8];
readData(menu);
Display(menu);
//system("pause");
return 0;
}
void header() {
char c= 61;
for (int i=0; i < 64; i++) {
cout << c;
}
cout << c << endl;
cout << endl;
cout << "Breakfast Menu" <<endl;
for (int i=0; i < 64; i++) {
cout << c;
}
cout << "" << c << endl;
cout << endl;
}
void readData(menuItemType item[]) {
int i=0;
ifstream in;
in.open("input.txt");
cout << fixed << setprecision(2);
while(!in.eof()) {
in >> item[i].itemName >> item[i].itemCost;
++i;
}
}
void Display(menuItemType item[]) {
int choice = 0, quantity = 0;
double total = 0.0, totalF = 0.0, tax = 0.0;
char exit = 'y';
int j = 1, z = 1, i = 1;
//the Menu
for (int i=0; i<8; i++){
cout << j << ". " << setw(18) << left << item[i].itemName << "$" << setw(10) << item[i].itemCost << endl;
j++;
}
cout << endl;
while(exit == 'y' || exit == 'Y') {
cout << "Please Enter your Selection or 0 to Exit : ";
cin >> choice;
if(cin.fail()) {
cout << "*******Invalid selection*******" << endl;
cin.clear();
cin.ignore(1000,'\n');
} else if (choice==0) {break; }
else {
cout<< "Enter Quantity: ";
cin>> quantity;
if (quantity==0) { break;}
else {
choice--;
total += (quantity * item[choice].itemCost);
tax = (total * .05);
totalF = total + tax;
cout << endl;
}
cout << endl;
cout << "======================================================" << endl;
cout << item[choice].itemName << "\t\t" << item[choice].itemCost << endl;
cout << "======================================================" << endl;
cout << "Do you want to continue (Y/N): ";
cin >> exit;
}
}
}
First off, you don't need a two dimensional array for this! You already have a one dimensional array of a suitable structure, as far as I can tell: Something which stores the name of the object and its price. What is somewhat missing is how many objects are currently in the array and how much space it has. If you want to go with the content of the entire array, make sure that you objects are correctly initialized, e.g., that the names are empty (this happens automatically, actually) and that the prices are zero (this does not).
I'm not sure if it is a copy&paste errors but the headers are incorrectly included. The include directives should look something like this:
#include <iostream>
The actual loop reading the values doesn't really work: You always need to check that the input was successful after you tried to read! Also, using eof() for checking that the loop ends is wrong (I don't know where people pick this up from; any book recommending the use of eof() for checking input loops is only useful for burning). The loop should look something like this:
while (i < ArraySize && in >> item[i].itemName >> item[i].itemCost)
++i;
}
This also fixes the potential boundary overrun in case there is more input than the array can consume. You might want to consider using a std::vector<Item> instead: this class keeps track of how many elements there are and you can append new elements as needed.
Note that you didn't quite say what you are stuck with: You'd need to come up with a clearer description of what your actual problem is. The above is just correcting existing errors and readjusting the direction to look into (i.e., forget about two dimensional arrays for now).