How to check integer overflow in Java?

Question

int a = 200000;

double b = (double) (a*a);

I would like to be able to calculate if (aa) (integers) is already too large (stackOverflow) to hold in the variable; in this case I know that multiplying (aa) leads to an int that will not fit -> the number will be wrong;

What I do not understand is how I can calculate this off the top of my head. How can I see if a value will not fit in an int variable??

Is there a way to do this without using methods, by just using the plain code and common sense? ;-)

Best regards,

Wouter

You can convert to long, do the calculation, and determine if the value is larger than Integer.MaxValue — Compass, Nov 09 '18 at 20:51
Too large for what? If `a` is a `long`, `float`, or `double`, your calculations will not overflow. Perhaps if your code has said `int a = 200000;`, it wouldn't be ambiguous. — Andreas, Nov 09 '18 at 21:00

score 4 · Answer 1 · edited May 31 '21 at 06:26

4

Two possible solutions.

Catch exception thrown after a call to the Math.…Exact methods: Math.multiplyExact, Math.addExact, Math.decrementExact, and so on.

 int a = 200000;

  try {
      int result = Math.multiplyExact(x, y);

  } catch(ArithmeticException e) {
      //if you get here, it is too big
  }

Or, check against the constants for minimum and maximum possible integer values.

 long test = (long)a*(long)a;
 if (test > Integer.MAX_VALUE || test < Integer.MIN_VALUE)
    // Overflow!

edited May 31 '21 at 06:26

Basil Bourque

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answered Nov 09 '18 at 20:57

Justin

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You only need to cast explicitly one operand to long, though. – Fran Marzoa Nov 09 '18 at 21:03

PumpkinBreath · Answer 2 · 2018-11-26T10:15:14.850

1

You could make use of the constant MAX_VALUE from the Integer wrapper class and cast to long instead (you will get a compile error if you cast to double but exceed it's range)

    int a = 200000;
    long b = (long) a * (long) a;
    if (b > Integer.MAX_VALUE) {
        // do something
    }

MAX_VALUE corresponds to the highest value an integer can be, (2^31)-1

edited Nov 26 '18 at 10:15

answered Nov 09 '18 at 20:58

PumpkinBreath

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I think this won't work, as the conversion to long is done after the int multiplication. – Fran Marzoa Nov 09 '18 at 21:04

How to check integer overflow in Java?

2 Answers2