3 * 1000000000 overflows as an int, but the variable is long long. Why?

Integer constants are, by default ints.

1000000000

That can fit into an int. So, this constant gets parsed as an int. But multiplying it by 3 overflows int.

10000000000

This is too big to an int, so this constant is a long long, so the resulting multiplication does not overflow.

Solution: explicitly use long long constants:

long long calcOne = 3 * 100000000LL;     // 3e8, essentially
long long calcTwo = 3 * 1000000000LL;    // 3e9, essentially
long long calcThree = 3 * 10000000000LL; // 3e10, essentially

What you do with a result doesn't affect how that result is calculated. So the fact that you store the result in a long long doesn't change the fact that the numbers you multiplied in the second line of code were not long longs and so they overflowed. In the third line of code, the constant is a long long, so the multiplication is performed on long longs.

The compiler saw this

long long calcOne = (int) 3 * (int) 100000000;     // 3e8, essentially
long long calcTwo = (int) 3 * (int) 1000000000;    // 3e9, essentially
long long calcThree = (int) 3 * (long long) 10000000000; // 3e10, essentially

And so the calcTwo right hand value was inferred as an int type and then over flowed. You see the over flow as a negative long.

long long calcOne = 3LL * 100000000LL;     // 3e8, essentially
long long calcTwo = 3LL * 1000000000LL;    // 3e9, essentially
long long calcThree = 3LL * 10000000000LL; // 3e10, essentially

To avoid this in the future, be explicit as to the types of your static values.To tell the compiler a number is a long long post fix it with LL.