Safely casting long to int in Java

What's the most idiomatic way in Java to verify that a cast from long to int does not lose any information?

This is my current implementation:

public static int safeLongToInt(long l) {
    int i = (int)l;
    if ((long)i != l) {
        throw new IllegalArgumentException(l + " cannot be cast to int without changing its value.");
    }
    return i;
}

Solution 1:

A new method has been added with Java 8 to do just that.

import static java.lang.Math.toIntExact;

long foo = 10L;
int bar = toIntExact(foo);

Will throw an ArithmeticException in case of overflow.

See: Math.toIntExact(long)

Several other overflow safe methods have been added to Java 8. They end with exact.

Examples:

• Math.incrementExact(long)
• Math.subtractExact(long, long)
• Math.decrementExact(long)
• Math.negateExact(long),
• Math.subtractExact(int, int)

Solution 2:

I think I'd do it as simply as:

public static int safeLongToInt(long l) {
    if (l < Integer.MIN_VALUE || l > Integer.MAX_VALUE) {
        throw new IllegalArgumentException
            (l + " cannot be cast to int without changing its value.");
    }
    return (int) l;
}

I think that expresses the intent more clearly than the repeated casting... but it's somewhat subjective.

Note of potential interest - in C# it would just be:

return checked ((int) l);

Solution 3:

With Google Guava's Ints class, your method can be changed to:

public static int safeLongToInt(long l) {
    return Ints.checkedCast(l);
}

From the linked docs:

checkedCast

public static int checkedCast(long value)

Returns the int value that is equal to value, if possible.

Parameters: value - any value in the range of the int type

Returns: the int value that equals value

Throws: IllegalArgumentException - if value is greater than Integer.MAX_VALUE or less than Integer.MIN_VALUE

Incidentally, you don't need the safeLongToInt wrapper, unless you want to leave it in place for changing out the functionality without extensive refactoring of course.

Solution 4:

With BigDecimal:

long aLong = ...;
int anInt = new BigDecimal(aLong).intValueExact(); // throws ArithmeticException
                                                   // if outside bounds

Solution 5:

here is a solution, in case you don't care about value in case it is bigger then needed ;)

public static int safeLongToInt(long l) {
    return (int) Math.max(Math.min(Integer.MAX_VALUE, l), Integer.MIN_VALUE);
}