Finding the sum of an array

[sarangheh510]

Could someone please explain to me why this the way as it is?

  #include<iostream>
using namespace std;

int number;//why it only seems to compile when initialized here.

int main()
{
    int sum=0;
    //had it initialized as int arr[]; and int number;
    //also had int arr[]={number};
    //could someone explain these differences to me?
    int arr[number];
    
    cout << "Please enter number." << endl; 
    cin >> number;
    cout << "Please enter " << number << "numbers" << endl;
    for(int i=1; i<number; i++)
    {
        cout << "Enter number" << i << endl;

    }
    sum += sum;
    cout << "Your sum is" << sum;
    return 0;
}

Please enter number.
Please enter 0numbers
Your sum is0

[SECMoG1T]

[codeint arr[number]][/code] is wrong number isn't initialized declaring an array especially a static
array would require you to provide its size during it's declaration you can als use a constant expression for declaration of the array size

int array [10]; /// an array of size 10
constexpr size_t  size=10;
int array [size]; ///an array of ten ints

If you want to request thr array size from the use then you can't use a static array
Use dynamic arrays instead

size_t size; 
cout <<"enter size"; cin>> size; 
int *arr= new int [size]; 

you will need to add values to your array after that ; calculate the sum but not as you have done above coz your sum+=sum is doing the wrong thing and is also outside the loop. then free the array storage

[sarangheh510]

Do I have to always set a [size] of an array or set the elements in the array when initializing an array? But in the case of the user inputting the size of an array, we first start off with putting a random number of array size for it, then changing it later in the code depending on the user input?

[sarangheh510]

Could someone modify my code properly?

[keskiverto]

There is something way more convenient than an array: http://www.cplusplus.com/reference/vector/vector/

However, you don't use the array anywhere. How could we know what it is for?

Please explain line 22 too.

[SECMoG1T]

When using static arrays it is only safe to intialize its size o declaration
we first start off with putting a random number of array size for it, then changing it later in the code  these is as dangerous as it sounds, you can never change the size of a static array it size must remain constant , the size is also part of its type hence changing it size might mean you are also changing the arry type

int arr [10]  // arr have the type int [10]  to 
//change the array size you will need to make another array of the new size and copy the
// elements of the older array into the new array
 which is costly since you cant delete the older array.

If you want to ddetermine the array size at runtime please use dynamic arrays or other containers such as vectors that are capable of efficient dynamic growth

[sarangheh510]

Woops I meant "sum += arr[i];" for line 22.

Could someone please modify this code to proper format with line 22 adjustments in mind!

[keskiverto]

No, that is your privilege.


How many times will the line 22 execute, when the program runs?

[sarangheh510]

I am confused. And it would just execute once after exiting the for loop, adding all the arr[i] up to make the sum.

[SECMoG1T]

#include<iostream> 
using namespace std; 

 int main() 
 { 
    int sum=0, size;
 
    cout << "Please enter the size of your array." << endl;
    cin >> size;

     int *array = new int [size];
     
     cout << "Please enter " << size << "integer values " << endl; 

      for(int i=0; i<size; i++)
          { 
              cout << "Enter number" << i << endl; 
              cin>> array [i];
           } 

       for (int i=0; i <size; i++)
            {
                sum += array [i];
            }

        cout << "Your sum is" << sum;
        delete array; 
         return 0; 
    }

[sarangheh510]

Can someone explain to me what line 11 does, "int *array = new int [size];"

I don't understand what the "*" does and the "new int" is.
@andy1992

[SECMoG1T]

int *array = new int [size]; is an int pointer , it points to a dynamic array of the size specified by the user. '*' show that the variable array is a pointer, * is called a dereference operator

Check here
http://www.cplusplus.com/doc/tutorial/pointers/
http://www.cplusplus.com/doc/tutorial/dynamic/
http://www.cplusplus.com/reference/new/operator%20new[]/