Conversion to Integers

Normally, a double value cannot be assigned to an int.

double x = 5.0;
// will not work
int y = x;

The reason for this is that there are numbers like 2.5, the infinities, and NaN which do not have an obvious way to be represented as an integer.

There are also numbers which a double can represent like 4207483647.0 and -9999999999.0 which happen to be too big or too small to fit into the limits of an int even though they do have an obvious "integer representation."

As such, to make an int out of a double you need to accept that it is a "narrowing conversion." The number you put in won't neccesarily be the number you get out.

To perform such a narrowing conversion, you need to put (int) before a literal or expression that evaluates to a double.

double x = 5.0;
// will be 5
int y = (int) x;

System.out.println(y);

Any decimal part of the number will be dropped. So numbers like 2.1, 2.5, and 2.9 will all be converted into simply 2.

int x = (int) 2.1;
int y = (int) 2.5;
int z = (int) 2.9;

System.out.println(x);
System.out.println(y);
System.out.println(z);

Any number that is too big to store in an int will be converted to the biggest possible int, 2³¹ - 1.

// 2147483647
System.out.println((int) 4207483647.0);

double positiveInfinity = 5.0 / 0.0;
// 2147483647
System.out.println((int) positiveInfinity);

Any number that is too small to store in an int will be converted to the smallest possible int, -2³¹.

// -2147483648
System.out.println((int) -9999999999.0);

double negativeInfinity = -5.0 / 0.0;
// -2147483648
System.out.println((int) negativeInfinity);

And NaN will be converted to zero.

double nan = 0.0 / 0.0;
System.out.println((int) nan);

When you use (int) to convert, we call that a "cast¹ expression". The (int) is a "cast operator." It even has a place in the operator precedence order just like +, -, ==, etc.

The main difference is that instead of appearing between two expressions like the + in 2 + 5, it appears to the left of a single expression.