3

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

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5
  • what do you mean by too large or too big Commented Nov 9, 2018 at 20:49
  • In Java plain code should be inside a method. Commented Nov 9, 2018 at 20:50
  • 1
    You can convert to long, do the calculation, and determine if the value is larger than Integer.MaxValue Commented Nov 9, 2018 at 20:51
  • 2
    Do you mean integer overflow instead of stack overflow? Commented Nov 9, 2018 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. Commented Nov 9, 2018 at 21:00

2 Answers 2

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5

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!
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1 Comment

You only need to cast explicitly one operand to long, though.
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

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1 Comment

I think this won't work, as the conversion to long is done after the int multiplication.

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