Thread: STL Vector problem with code

Code:

    for (forLoopNumber = 0; forLoopNumber <= numberOfElements; forLoopNumber++){
        sum += array[forLoopNumber];
    }
    cout<<"Average equals: " << sum/numberOfElements;

First of all, your for condition doesn't look quite right.

Second, the expression sum/numberOfElements is of type int, meaning that any fractional part of the quotient will be discarded (number will be rounded down).

ok I belive I fixed the problem

somthing in the for loop wasn't working out... I also changed those that needed to be changed to float

Fixed code:

Code:

#include<iostream>
#include<vector>

using namespace std;

int main()
{
    float numberOfElements;
    vector<int> array;
    float sum=0;
    int arrayNumber = 0;
    int loopNumber = 0;
    bool finished = false;
    
    cout<< "How many numbers will you be entering?"; 
    cin>> numberOfElements;
    cin.ignore();
    
    array.push_back(1);
    cout<<"\n Enter the numbers:\n\t\t\t";
    while(finished == false){
         cin>> array[arrayNumber];
         cin.ignore();
         if (arrayNumber + 1== numberOfElements){
         finished = true;
         }
         if (arrayNumber + 1 != numberOfElements){
         arrayNumber++;
         array.push_back(1);
         }
    }
    while (loopNumber < numberOfElements){
        sum += array[loopNumber];
        loopNumber++;
    }
    cout<<"Average equals: " << sum/numberOfElements;
    cin.get();
}

Some optimization in the code:

Code:

#include<iostream>
#include<vector>

using namespace std;

int main()
{
    float numberOfElements;
    vector<int> array;
    float sum=0;
    int arrayNumber = 0;
    int loopNumber = 0;
    bool finished = false;
    
    cout<< "How many numbers will you be entering?"; 
    cin>> numberOfElements;
    cin.ignore();
    
    array.push_back(1);
    cout<<"\n Enter the numbers:\n\t\t\t";
    
    while(finished == false){
         cin>> array[arrayNumber];
         cin.ignore();
         
         
         if (arrayNumber + 1== numberOfElements){
              finished = true;
         }
         
         else {
              arrayNumber++;
              array.push_back(1);
         }
    }
    
    for (loopNumber = 0; loopNumber < numberOfElements; loopNumber++){
        sum += array[loopNumber];
    }
    
    cout<<"Average equals: " << sum/numberOfElements;
    cin.get();
}

You are completely mistreating the vector. The whole point of the vector is that you do not need to worry about size. Read into a temp variable, then push that back. Don't push back an arbitrary number and read directly into the vector.
You basically got multiple if statements in the first loop. You can easily replace the second one with "else". Less typing, less prone to bugs and less code duplication. Better for all.
And the last loop should really be a for loop.
You've also borked the indentation in some places, especially in your first loop and the if statements.

Good training! Then you won't make the same mistake next time

Originally Posted by jamort

Code:

    float numberOfElements;
    // ...
    cout<< "How many numbers will you be entering?"; 
    cin>> numberOfElements;

Hmm so if I would like to enter 2.3 elements that's okay?
Nope, aparently 2.3 means an infinite number of items....

Don't trust the user.
Instead, you can do:

int temp;
cin >> temp;
array.push_back(temp);

Don't think I did what you said to do right.... let me know... I edited this because the way i had if there was a 0 or negative number entered it would mess up

Code:

#include<iostream>
#include<vector>

using namespace std;

int main()
{
    int numberOfElements = 0;
    int arrayNumber = 0;
    int loopNumber = 0;
    vector<int> array;
    float sum=0;
    bool finished = false;
    while (numberOfElements <= 0){
          cout<< "How many numbers will you be entering?" << endl; 
          cin>> numberOfElements;
          cin.ignore();
          
          if (numberOfElements <= 0){
             cout<< endl << "Error... Please enter a positive integer"<<endl;;
          }
    }
    
    cout<<" Enter the numbers:\n\t\t\t";
    
    array.push_back(numberOfElements);
    
    while(finished == false){
         cin>> array[arrayNumber];
         cin.ignore();
         
         
         if (arrayNumber + 1== numberOfElements){
              finished = true;
         }
         
         else {
              arrayNumber++;
         }
    }
    
    for (loopNumber = 0; loopNumber < numberOfElements; loopNumber++){
        sum += array[loopNumber];
    }
    
    cout<<"Average equals: " << sum/numberOfElements;
    cin.get();
}

Code:

#include<iostream>
#include<vector>

using namespace std;

int main()
{
    int numberOfElements = 0;
    int arrayNumber = 0;
    int loopNumber = 0;
    vector<int> array;
    float sum=0;
    bool finished = false;
    int temp;
    while (numberOfElements <= 0){
          cout<< "How many numbers will you be entering?" << endl; 
          cin>> numberOfElements;
          cin.ignore();
          
          if (numberOfElements <= 0){
             cout<< endl << "Error... Please enter a positive integer"<<endl;;
          }
    }
    
    cout<<" Enter the numbers:\n\t\t\t";
    
//    array.push_back(numberOfElements);
    
    while(finished == false){
         cin >> temp;
        array.push_back(temp);
         cin.ignore();
         
         
         if (arrayNumber + 1== numberOfElements){
              finished = true;
         }
         
         else {
              arrayNumber++;
         }
    }
    
    for (loopNumber = 0; loopNumber < numberOfElements; loopNumber++){
        sum += array[loopNumber];
    }
    
    cout<<"Average equals: " << sum/numberOfElements;
    cin.get();
}

Originally Posted by jamort

ok... I see now so when you pushback an array the number that is passed becomes the new value...

I'm not sure what you mean by that, but basically, push_back takes an argument and adds that argument to the end of its internal array. A vector is basically a dynamic array.
Safe and easy. You don't have to worry about a thing.

Yes, it will. It works with any type.