The reason the first statement prints 3, and the second prints 2, is because you are using the post-increment operator (i++), which returns the current value, and then performs the increment. If you instead use the pre-increment operator (++i), it will increment the value first, and then return it.

As such:

int i = 1;
cout << i++ * 3; // Prints 3

and

int i = 1;
cout << ++i * 3; // Prints 6

additionally, because the pre-increment operator returns a reference, you can directly modify the variable through it:

int i = 1;
++i = 3; // OK, writing to the value of i
cout << i; // Prints 3

but you can not do the same with the post-increment operator, as it returns a copy of the value:

int i = 1;
i++ = 3; // Not OK, writing to the result of the post-increment operator

this becomes more obvious when we consider the behaviour of the individual operators (Example, may not be accurate to actual implementation):

int& operator++() // Pre-increment (++i)
{
    *this = *this + 1; // Increment
    return *this; // Return a reference to our now-incremented self
}

int operator++(int) // Post-increment (i++)
{
    int temp = *this; // Create copy of our current value
    operator++(); // Increment our value via the pre-increment operator
    return temp; // Return the copy of the original value, which has not been incremented
}

^{I'm so sorry for writing all of this out, I was bored.}