Thread: Can anyone help me?

Well, you can either read up on dynamic memory so that you know how to allocate memory for your array, or you can just use the vector class.

Code:

while(TRUE)
{
  cout<<"Enter a number: ";
  cin >> num;
  if(num == SENTINEL) break;
  myVector.pushback(num);
}

I think that'll do what you want.

> #include <iostream.h>

Code:

#include <iostream>

>

Code:

using namespace std;

or

Code:

using std :: cout;
using std :: cin; // etc.

or

Code:

std :: cout << "..." << std :: endl;

>void main()

Code:

int main()

- SirCrono6

P.S. I'm a standard boy =/

example output;

enter number (one at a time)
5
6
7
-1

enter second number (one at time)

6
4
3
2
-1

number enter where 567 and 6432
the sum is 6999

What I have below is what I have now but the if statement doesn't really do what I want it to. It end the number but then it messes up the sum part. Any ideas???

Code:

     
      
 
     
 #include<iostream.h>

 int convertNumber(int [], int);
 void showvalues(int[]);
 
 void main ()
 {
	int num1[5], num2[5];
	int val1, val2;
			
	cout<< "enter number" <<endl;
	for (int count = 0; count < 5; count++)
	{
		cin >> num1[count];
		if (num1[count] == -1)
			break;
	}
	  
	val1 = convertNumber(num1, 5);
	cout << endl;
		

	cout<< "enter number" <<endl;
	for (int count1 = 0; count1 < 5; count1++)
	{
		cin >> num2[count1];
		if (num2[count1] == -1)
			break;
	}
	
	val2 = convertNumber(num2, 5);
	cout << endl;
	
	cout << "You entered the numbers:" << val1 << " and " << val2 << endl;	 
	cout << "the sum of the numbers is: " <<  val1 + val2 << endl;

 }


int convertNumber(int nums[], int arySize)
{
    int multiplier = 1;
    int val = 0;
    for (int x = arySize-1 ; x >= 0; x--)
    {
        val += nums[x] * multiplier;
        multiplier *= 10;
    }
	return val;
}

Originally posted by sflyers

Code:

int convertNumber(int nums[], int arySize)
{
    int multiplier = 1;
    int val = 0;
    for (int x = arySize-1 ; x >= 0; x--)
    {
        val += nums[x] * multiplier;
        multiplier *= 10;
    }
	return val;
}

Since the last value entered is -1, how about

Code:

int convertNumber(int nums[])
{
    int val = 0;
    int j = 0;
    while (nums[j] > -1)
    {
        val = (val * 10) + nums[j];
        j++;
    }
    return val;
}

This way you can have a larger number if you need it (and you use long)