Safely casting long to int in Java

Question

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;
}

Answer 1

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);
}

Answer 2

One other solution can be:

public int longToInt(Long longVariable)
{
    try { 
            return Integer.valueOf(longVariable.toString()); 
        } catch(IllegalArgumentException e) { 
               Log.e(e.printstackstrace()); 
        }
}

I have tried this for cases where the client is doing a POST and the server DB understands only Integers while the client has a Long.

Answer 3

With BigDecimal:

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

Answer 4

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.

Answer 5

Java integer types are represented as signed. With an input between 2³¹ and 2³² (or -2³¹ and -2³²) the cast would succeed but your test would fail.

What to check for is whether all of the high bits of the long are all the same:

public static final long LONG_HIGH_BITS = 0xFFFFFFFF80000000L;
public static int safeLongToInt(long l) {
    if ((l & LONG_HIGH_BITS) == 0 || (l & LONG_HIGH_BITS) == LONG_HIGH_BITS) {
        return (int) l;
    } else {
        throw new IllegalArgumentException("...");
    }
}

Answer 6

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);