Hault!

Lets suppose I have an integer called number. Would it be possible to set boundaries on that integer so it is not possible to input a number larger than say 30000. I don't mean after you press enter, it runs some code to check if number>30000. I mean, at runtime, before you press enter...is it possible to physically not let the person enter a number higher than 30000, or if not, is it possible to not allow them to enter a number over so many digits?

Not really... unless you use a non-standard way to control each keypress of the user.

I guess you'd have to write your own input function then. This is one I've made (uses conio.h):
Specify 0 as the parameter if you don't wish to have an upper limit. This might be useful too since it only accepts digits (not letters/other characters).

Code:

#include <iostream>
#include <conio.h>

using namespace std;


#define MAXDIGITS 9


int ReadNumber(int MaxNumber)
{
	//Data
	int i;
	char Key = 0;
	char Buffer[MAXDIGITS] = {'0'};
	int CurrentPointer = 0;
	int Result = 0;

	//Loop until ENTER is pressed
	while((Key != (char)13) || (CurrentPointer == 0))
	{
		//Get the next keypress
		Key = getch();

		//The user pressed a digit
		if((Key >= '0') && (Key <= '9') && (CurrentPointer < (MAXDIGITS - 1)))
		{
			//Accept the input only if it doesn't exceed the limits
			if((MaxNumber == 0) || (((Result * 10) + (int)(Key - '0')) <= MaxNumber))
			{
				Buffer[CurrentPointer] = Key;
				CurrentPointer++;
				cout << Key;
			}
		}
		//The user pressed BACKSPACE
		else if((Key == '\b') && (CurrentPointer > 0))
		{
			Buffer[CurrentPointer] = '0';
			CurrentPointer--;
			cout << '\b';
			cout << ' ';
			cout << '\b';
		}

		//Calculate the result
		Result = 0;
		for(i = 0; i <= (CurrentPointer - 1); i++)
		{
			Result *= 10;
			Result += (int)(Buffer[i] - '0');
		}
	}

	//Return the number
	return Result;
}


//Test
int main()
{
	int X;

	cout << "Enter a number: ";
	X = ReadNumber(500);
	cout << endl << "Your input was: " << X << endl;

	return 0;
}