Array size..!!

[Kobe24]

I am asking the user to input only 2 integers and I want to check if the user did input 2 integers using an array. Here is my code. Thanks for the help

  #include <iostream>
#include <string>

using namespace std;


int main()
{
  int a(0);
  int SIZE = 2;
  int array[SIZE];
  cout <<"Enter 2 positive integers: ";
  cin >> a;
  if(a > 0)
    {
      if(a.size() >  array[SIZE])
	{
	  cout << "Too many numbers" << endl;
	}
	 else
	   {
	     cout << a;
	   }
     }

    return 0;
}

[Gaminic]

'a' is an int, it doesn't have a size() function, because it is always only one element.

array[SIZE] isn't your size; SIZE is your size. array[SIZE] is the SIZE'th element of array, which doesn't exist, because array has SIZE elements going from [0, SIZE-1].

On line 13, you're telling the program to read in *one* int. Anything beyond that is ignored (sort of). You can tell it to read exactly two ints quite similarly:

int a;
int b;
cin >> a >> b;

This will read in (the first) two ints and save them into a and b respectively. If you want to read a variable number of ints, you can use a loop.

[Kobe24]

What if I am only allowed to use one variable aka int a. Do I do it something like this?

for(int i = 0 ; i < (array size of the integer entered) ; i++)
{

  etc...
}

[3qX21hU5]

What exactly are you trying to do?

[Kobe24]

I am trying to store as many numbers in the array and check if there is only 2 integers. I know it sounds stupid to use an array but I am just practicing to learn how to use size of array for other different things like odd/even/negative/non-negative..etc..!!

[3qX21hU5]

Well if you want to store the number into a array directly you could use a loop like this

// Use size_t when making a index for a array. Also SIZE is the array's size.
for (size_t i = 0; i != SIZE; ++i)
{
    cin >> myArray[i];
    ++numberOfInputs;
}

That will read the users input directly into the arrays elements. Then if you wanted to keep track of how many times the user inputted numbers you could make a counter that keeps track of it. If that is what you are asking, if not let me know.

[Gaminic]

That won't work, because it'll stop at SIZE and therefore never know if it's bigger than SIZE.

[3qX21hU5]

Huh how could the array be bigger then SIZE? and it stops 1 before SIZE.

Anyways I am not sure I understand what the OP is trying to do to. I don't really get the part of storing as many number in the array and the checking if there is only 2 integers parts.

Maybe that is why my example wont work? Otherwise it does what I believed he wanted just fine (Grab user input and put that input into the array), though now that I look at it again there is no point for the counter since it will always equal SIZE - 1 since there is no other way to break out of the loop.

[Chervil]

I'm not sure that I understand the question. But I would use a string and getline. Then use a stringstream to extract as many integers as possible from that string.

[3qX21hU5]

Lol I think all of us have different ideas on what he is trying to do. Kobe would you elaborate a bit more on what you are trying to do?

[AkramIzzeldin]

i guess you want the input to contain 2 digits only no less or more not 2 ints if that's what you mean here is how:
any number contains 2 digits only should be between the values 10-99 so you can test the input to be larger than 9 and smaller than 100.
I hope this is what you want.

[Minimacfox]

Not sure if you meant 2 individual values or just to check if you entered 2 integers in one value at a time. So I did both.

main(){
int amount = 0;
int i = 0;
const int SIZE = 2;
int my_array[SIZE];
    while(amount < 2){
        cout << "Enter integer " << (i+1) << ": ";
        (cin >> my_array[i]).get();
        if(!(my_array[i] > 9 && my_array[i] < 100)){
            system("cls");
            cout << "You didn't input 2 integers as a value!\nTry again!\n";
            system("pause");
            system("cls");
            continue;
        }
    i++;
    amount++;
    }
}