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

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)

Answer 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);

Answer 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.

Answer 4

With BigDecimal:

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

Answer 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);
}

Answer 6

DONT: This is not a solution!

My first approach was:

public int longToInt(long theLongOne) {
  return Long.valueOf(theLongOne).intValue();
}

But that merely just casts the long to an int, potentially creating new Long instances or retrieving them from the Long pool.

---

The drawbacks

1. Long.valueOf creates a new Long instance if the number is not within Long 's pool range [-128, 127].

2. The intValue implementation does nothing more than:

    return (int)value; 

So this can be considered even worse than just casting the long to int .

Answer 7

I claim that the obvious way to see whether casting a value changed the value would be to cast and check the result. I would, however, remove the unnecessary cast when comparing. I'm also not too keen on one letter variable names (exception x and y , but not when they mean row and column (sometimes respectively)).

public static int intValue(long value) {
    int valueInt = (int)value;
    if (valueInt != value) {
        throw new IllegalArgumentException(
            "The long value "+value+" is not within range of the int type"
        );
    }
    return valueInt;
}

However, really I would want to avoid this conversion if at all possible. Obviously sometimes it's not possible, but in those cases IllegalArgumentException is almost certainly the wrong exception to be throwing as far as client code is concerned.

Answer 8

Java integer types are represented as signed. With an input between 2 ³¹ and 2 ³² (or -2 ³¹ and -2 ³² ) 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 9

(int) (longType + 0)

但龙不能超过最大值:)

Answer 10

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.