My question is to record the user 5 inputs and matching them to the correct code. singleperson code is 1, couple code is 2 and family is 3. Hence the users input will only be these 3 numbers.At the end I must calculate the total of the total number of people by their category. I ran the code and got runtime error saying stack around variable "groups" is corrupted and always have only 5 singles while the couples and families are 0. Sorry, I just started school and I'm a bit blurry..
int singleperson=0;
int couple=0;
int family3=0;
int groups[]={0,0,0,0,0};
cout << "Enter group #1:";
cin >>groups[0];
cout << "Enter group #2:";
cin >>groups[1];
cout << "Enter group #3:";
cin >>groups[2];
cout << "Enter group #4:";
cin >>groups[3];
cout << "Enter group #5:";
cin >>groups[4];
for (int a=0; a<=4;a++)
{
if(groups[a]=1)
{
singleperson= singleperson + 1;
}
else if(groups[a]=2)
{
couple = couple +1;
}
else
{
family3= family3+1;
}
}
cout<<"Statistics"<<endl;
cout <<singleperson<<"Singles"<<endl;
cout<<couple<<"Couples"<<endl;
cout<<family3<<"Families"<<endl;
Because size of group is 4 and you are trying to insert into 5th place. And as pointed by #antonio-garrido to check equality we use == and = is used for assignment.
int groups[]={0,0,0,0,0};
for (int a=0 ; a <= 4; a++) {
if(groups[a] == 1) {
singleperson= singleperson + 1;
} else if(groups[a] == 2) {
couple = couple +1;
} else {
family3= family3+1;
}
}
Related
so im new to c++ and im doing a program that takes user inputs for any amount number say 5 so i will get 5 inputs from user and calculate the sum of it ,i did make the program but what i want for the output is say
"Enter Input 1:xx
"Enter Input 2:xx
so on and on as the user input say 5 so it goes on for 5 times however my program takes the user input and i enter it, it dosnt say enter input 1 ,so i want to show the enter input 1 enter 2 part hope someone can help me with this sorry for my poor explanation
#include<iostream>
using namespace std;
int main() {
while (true) {
// prompts the user to ask how many inputs they want
int x;
cout << "Enter input : ";
cin >> x;
// If x = -1 dont repeat the loop
if (x == -1)
break;
// get the input from above and calculate the total of the input
int sum = 0;
for (int i = 0; i < x; i++) {
int value;
cin >> value;
sum += value;
}
// Output the total
cout << "Output total: " << sum << endl;
}
system("pause");
return 0;
}
Okay to start out, there are times when to use break and there are times not to. This scenario is not meant for them. Although, I used one in my code I did it because I am rushing. I would recommend to take the code snippet, learn from it, and see how to optimize it :)
Also, just for future purposes its important to understand your task at hand and be able to communicate it well so others can help debug and answer your question.
Heres what I think your question is:
"I want to make a program in cpp that allows the user to first submit how many numbers they would like to input. From there I would then ask them for the indicated number of inputs. After gathering all inputs I will then add those numbers together and output the sum. If at any point they decide to type in -1, I will stop asking for inputs and give their sum on the spot."
#include<iostream>
using namespace std;
int main() {
bool runProgram = true;
cout << "Hi welcome to my sum calculator program!\n";
cout << "This program will prompt you for a number of inputs and then calculate the total of them.\n";
cout << "If you no longer want to be prompted for numbers at any time type in -1!\n";
cout << "Press enter to begin!\n";
cin.get();
while (runProgram) {
// prompts the user to ask how many inputs they want
int x;
cout << "How many inputs?\n";
cin >> x;
// If x = -1 dont repeat the loop
if (x == -1){
runProgram = false;
}else{
// get the input from above and calculate the total of the input
int sum = 0;
int val = 0;
for (int i = 0; i < x; i++) {
cout<< "Input #" << i+1;
cin >> val;
if(val == -1){
runProgram = false;
break;
}
sum += val;
}
// Output the total
cout << "Your Output total is " << sum << endl;
}
}
system("pause");
return 0;
}
I'm beginning with C++. The question is: to write a program to input 20 natural numbers and output the total number of odd numbers inputted using while loop.
Although the logic behind this is quite simple, i.e. to check whether the number is divisible by 2 or not. If no, then it is an odd number.
But, what bothers me is, do I have to specifically assign 20 variables for the user to input 20 numbers?
So, instead of writing cin>>a>>b>>c>>d>>.. 20 variables, can something be done to reduce all this calling of 20 variables, and in cases like accepting 50 numbers?
Q. Count total no of odd integer.
A.
#include <iostream>
using namespace std;
int main(int argc, char** argv)
{
int n,odd=0;
cout<<"Number of input's\n";
cin>>n;
while(n-->0)
{
int y;
cin>>y;
if(y &1)
{
odd+=1;
}
}
cout<<"Odd numbers are "<<odd;
return 0;
}
You can process the input number one by one.
int i = 0; // variable for loop control
int num_of_odds = 0; // variable for output
while (i < 20) {
int a;
cin >> a;
if (a % 2 == 1) num_of_odds++;
i++;
}
cout << "there are " << num_of_odds << " odd number(s)." << endl;
If you do really want to save all the input numbers, you can use an array.
int i = 0; // variable for loop control
int a[20]; // array to store all the numbers
int num_of_odds = 0; // variable for output
while (i < 20) {
cin >> a[i];
i++;
}
i = 0;
while (i < 20) {
if (a[i] % 2 == 1) num_of_odds++;
i++;
}
cout << "there are " << num_of_odds << " odd number(s)." << endl;
Actually, you can also combine the two while-loop just like the first example.
Take one input and then process it and then after take another intput and so on.
int n= 20; // number of input
int oddnum= 0; //number of odd number
int input;
for (int i = 0; i < n; i ++){
cin >> input;
if (input % 2 == 1) oddnum++;
}
cout << "Number of odd numbers :"<<oddnum << "\n";
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 8 years ago.
Improve this question
I'm new to C++ and trying to create a lottery game for a college project.
I have a for loop to check that there are no duplicate numbers in the array entered. This works absolutely fine when you take out the section of code to produce the random numbers.
As soon as I add the random number section back in, the for loop just gets stuck. It will continuously tell me that i have already entered the number when its trying to store the first number.
I have attached all of my code, apologies if you don't need it all.
#include <iostream>
#include <stdio.h>
#include <time.h>
#include <stdlib.h>
using namespace std;
//int loto[6];
int main()
{
int numbers[6];
//void rand_num(int loto[6], int count);
int loto[6]; //used to store the loto numbers
//int james = 0;
//int l,j; //used in checking any duplicated
srand(time(0));
for(int count=0; count<6; count++)
{
loto[count] = (rand()%49)+1;
cout << loto[count] << endl;
}
//declares the variable i to increase each time a number is entered.
//this will only go as high as 6
for(int i=0;i<6;i++)
{
cout<<" " << i<<" : Please enter your lottery numbers: "<<endl;
cin>>numbers[i];
if ((numbers[i] >= 50) | (numbers[i] == 0))
do
{
{
//checks to see if the first number entered is above 50 or = to 0 and rejects it
cout << "The Number must be between 1-49, please select again. " << endl;
cin >> numbers[i];
}
}
while ((numbers[i] >= 50) | (numbers[i] == 0));
//----------------------------------------------------------------------------
//this section of code is a loop within a loop to check the number entered against all numbers already stored.
//makes l the same as i effectively
for(int l=0;l<6;l++)
{
//makes j one more than l
for(int j=l+1;j<7;j++)
{
if( numbers[l] == numbers[j] )
do
{
{
cout << "Number has already been chosen, please re-enter number " << endl;
cout << endl;
cin >>numbers[i];
//checks the number that is re-entered is not <50 or = 0
//if so it rejects it and asks for another as above.
if ((numbers[i] >= 50) | (numbers[i] == 0))
do
{
{
cout << "The Number must be between 1-49, please select again. " << endl;
cin >> numbers[i];
}
}
while ((numbers[i] >= 50) | (numbers[i] == 0));
}
}
while (numbers[l] == numbers[j]);
}
}
}
//----------------------------------------------------------------------------
//this displays the numbers that have been chosen.
cout << "Your Numbers are: " << endl;
for (int i = 0; i<6; i++)
{
cout << " " << numbers[i];
}
return 0;
}
I'm not sure this is the real problem but it is a bug. Try to correct it and see if it helps.
for(int l=0;l<6;l++)
{
//makes j one more than l
for(int j=l+1;j<7;j++)
{
if( numbers[l] == numbers[j] )
The inner-loop will reach j==6 so you will access outside the array. The outer-loop shall have 5 as the limit and the inner-loop shall have 6 as the limit.
EDIT:
After looking a bit more at your code I can see that you are using numbers[] without initializing it. The two nested for-loops will compare all elements in numbers. But if the user have only entered 2 numbers, the rest is unitialized and can give unintended results.
Further - you don't need to check all elements againt all elements every time. Just check the newly entered number (index by i) with all previous numbers.
Finally you will probably need something like:
if (!(cin >> numbers[i])) {
cout << "Please enter numbers only." << endl;
cin.clear();
cin.ignore(10000,'\n');
}
to handle input not being integer, e.g. "text"
And to minor things:
You should also check for negative numbers.
You are using | instead of ||. It will work fine but || seems more correct as it is the logical OR (while | is a binary OR).
I present to you all a program I'm working on for my college programming course. I still have a little ways to go before it completely meets my assignment's requirements, but I've gotten a basic draft of the program error-free (supposedly) and it appears to run… but then it suddenly kicks me into Xcode's debugger and gives me:
Thread 1: EXC_BAD_ACCESS(code=2, address=0x7fff95c1e5f5)
Here's the command line output, up until it kicks me out:
-----------------------
Quarterly_sales_taxator
-----------------------
How many company divisions will we be dealing with? 2
Am I correct in assuming that there are 4 sales quarters? yes
Please enter the sales Company Division #1 brought in for Sales Quarter #1 20
(lldb)
Here's my code:
//
// quarterly_sales_taxator.cpp
// Ch. 7 program #7
//
// Created by John Doe on 11/27/12.
//
#include <iostream>
#include <iomanip>
#include <string>
#include <sstream>
#include <cctype>
using namespace std;
void read_company_divisions_and_sales_quarters(double **, int, int);
//void write_company_divisions_and_sales_quarters_to_array(double **, int, int); // This will be used later on to read data from a file.
void display_quarterly_sales_array(double **, int, int);
string temp; // A global temporary placeholder variable; I use this several times.
int main()
{
int COMPANY_DIVISIONS,
SALES_QUARTERS = 4;
double **quarterly_sales_form;
cout << "\n\n-----------------------\nQuarterly_sales_taxator\n-----------------------\n\n";
cout << "\nHow many company divisions will we be dealing with? ";
getline(cin, temp);
stringstream(temp)>>COMPANY_DIVISIONS;
while (COMPANY_DIVISIONS < 1 || isdigit(COMPANY_DIVISIONS == false))
{
cout << "\n\n------"
<< "\nError:"
<< "\n------"
<< "\n\nYou have entered an invalid choice."
<< "\nPlease type a number greater than zero. ";
getline(cin, temp);
stringstream(temp)>>COMPANY_DIVISIONS;
}
cout << "\n\nAm I correct in assuming that there are 4 sales quarters? ";
getline(cin, temp);
// Convert to uppercase.
for (int count = 0; count < temp.length(); count ++)
{
temp[count] = toupper(temp[count]);
}
if (temp == "NO" || temp == "NOPE" || temp == "INCORRECT" || temp == "YOU ARE NOT" || temp == "YOU ARE INCORRECT" || temp == "NEGATIVE" || temp == "NEGATORY")
{
cout << "\nOk, then how many sales quarters are we dealing with? ";
getline(cin, temp);
stringstream(temp)>>SALES_QUARTERS;
}
cout << endl << endl;
// This sets up the 2d array.
quarterly_sales_form = new double *[COMPANY_DIVISIONS];
for (int count = 0; count < COMPANY_DIVISIONS; count ++)
{ quarterly_sales_form[COMPANY_DIVISIONS] = new double [SALES_QUARTERS]; }
read_company_divisions_and_sales_quarters(quarterly_sales_form, COMPANY_DIVISIONS, SALES_QUARTERS);
// write_company_divisions_and_sales_quarters_to_array(quarterly_sales_form, COMPANY_DIVISIONS, SALES_QUARTERS); // I'll add this feature later.
cout << "\n\nHere's what you entered:\n\n";
display_quarterly_sales_array(quarterly_sales_form, COMPANY_DIVISIONS, SALES_QUARTERS);
// Since we used a series of pointers, we need to free the allocated space back up.
for (int count = 0; count < COMPANY_DIVISIONS; count ++)
{ delete[] quarterly_sales_form[COMPANY_DIVISIONS]; }
delete[] quarterly_sales_form;
return 0;
}
/*############################################
# read_company_divisions_and_sales_quarters #
############################################*/
void read_company_divisions_and_sales_quarters(double **array, int DIVISIONS, int QUARTERS)
{
for (int count = 0; count < QUARTERS; count++)
{
for (int index = 0; index < DIVISIONS; index++)
{
cout << "\nPlease enter the sales Company Division #" << count+1 << " brought in for Sales Quarter #" << index+1 << " ";
getline(cin, temp);
stringstream(temp) >> array[count][index];
}
}
}
/*################################
# display_quarterly_sales_array #
#################################*/
void display_quarterly_sales_array(double **array, int DIVISIONS, int QUARTERS)
{
for (int count = 0; count < DIVISIONS; count++)
{
cout << "\nCompany division #" << count+1 << ":\n";
for (int index = 0; index < QUARTERS; index++)
{ cout << array[count][index] << ", "; }
}
}
Can some kind soul please tell me what I'm doing wrong?
{ quarterly_sales_form[COMPANY_DIVISIONS] = new double [SALES_QUARTERS]; }
In this line, COMPANY_DIVISIONS should be count.
In addition to what Dan Hulme said, it seems this line
stringstream(temp) >> array[count][index];
should really be
std::istringstream(temp) >> std::skipws >> array[index][count];
In addition to using std::istringstream rather than std::stringstream and making sure that an lvalue is at hand, which isn't strictly needed until the type read becomes more interesting, this also reverses the indices: index runs over COMPANY_DIVISIONS and count over SALES_QUARTERS.
The real question is, of course: Who hands out assignments like this? Pointer manipulations and allocations are best left to low-level library writers. This is C++ not C: we can and should use abstractions. Getting this code exception safe is a major challenge and there is no point in teaching people how to write broken (e.g. exception unsafe) code.
Sorry about last time for those who saw my previous thread. It was riddled with careless errors and typos. This is my assignment:
"Write a program that will enable the user to enter a series of non-negative numbers via an input statement. At the end of the input process, the program will display: the number of odd numbers and their average; the number of even numbers and their average; the total number of numbers entered. Enable the input process to stop by entering a negative value. Make sure that the user is advised of this ending condition."
And here is my code:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int number, total1=0, total2=0, count1=0, count2=0;
do
{
cout << "Please enter a number. The program will add up the odd and even ones separately, and average them: ";
cin >> number;
if(number % 2 == 0)
{
count1++;
total1+=number;
}
else if (number >= 0)
{
count2++;
total2+=number;
}
}
while (number>=0);
int avg1 = total1/count1;
int avg2 = total2/count2;
cout << "The average of your odd numbers are: " << avg1 << endl;
cout << "The average of your even numbers are " << avg2 << endl;
}
It seems to be working fine, but when I enter a negative number to terminate the program, it includes it with the rest of the averaged numbers. Any advice to get around this? I know it's possible, but the idea escapes me.
Your main loop should be like this:
#include <iostream>
for (int n; std::cout << "Enter a number: " && std::cin >> n && n >= 0; )
{
// process n
}
Or, if you want to emit a diagnostic:
for (int n; ; )
{
std::cout << "Enter a number: ";
if (!(std::cin >> n)) { std::cout << "Goodbye!\n"; break; }
if (n < 0) { std::cout << "Non-positve number!\n"; break; }
// process n
}
After here:
cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
cin >> number;
Immediately check if the number is negative
if(number < 0) break;
Now you wouldn't need to use your do-while loop in checking if the number is negative. Thus, you can use an infinite loop:
while(true) {
cout << "Please enter a number. The program will add up the odd and even ones seperately, and average them: ";
cin >> number;
if(number < 0) break;
// The rest of the code...
}
ADDITIONAL:
There is something wrong in your code. You aren't showing the user how much the number of even and odd numbers are, and the total number of numbers entered.
ANOTHER ADDITIONAL: You should use more meaningful variable names:
int totalNumEntered = 0, sumEven = 0, sumOdd = 0, numEven = 0, numOdd = 0;
Of course I am not limiting you to these names. You can also use other similar names.
FOR THE INTEGER DIVISION PROBLEM:
You must cast your expression values to the proper type (in this case, it is float). You should also change the averages variables' types to float:
float avg1 = float(total1) / float(count1);
float avg2 = float(total2) / float(count2);
Immediately after cin >> number, check for < 0, and break if so. Try to step through the program line by line to get a feel for the flow of execution. Have fun learning, and good luck!