If bit 0 is set, then it is an odd number. Only even numbers are divisible by 2.
If you and a0 with 1, and store the result in v0, v0 will contain 1 when a0 is NOT divisible by 2.
You are doing it right, but it can be done with 1 instruction.

Ctrl+F5(Start without debuging)

void addition(float a, float b, float total)
{
	cout << "Enter two integers (separated by a single space): ";
	cin >> a, b;
	total = a + b;
	cout << "The sum is: " << total << endl;
}

If you have declared your add function like this, you have to call it like this.

addition(firstArgumentToBePassed, secondArgumentToBePassed, etc, etc...);

As you call it now, you could just change you functions.

void addition()
{
        int a, b, total = 0;
	cout << "Enter two integers (separated by a single space): ";
	cin >> a, b;
	total = a + b;
	cout << "The sum is: " << total << endl;
}

Don't write "Enter two integers", and then use floats BTW.

You can declare b before, but you can't assign a to it before anything is assigned to a.

int a;

int b;

printf("Enter a number:");
scanf("%d",&a);

b = a+1;

printf("%d", B)/>/>; 

It is the same for all languages I think, anything else would be strange.

Well, in C/++ you COULD, but it would be a garbage value.

http://www.dreaminco...-state-manager/
This should do it for a 2D game, since it is for a 2D game. Good tutorial also.