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The output for a Java program I wrote is giving me "Infinity" when I should be getting a number. I'm using floats, and I don't think my numbers are going out of the range of what floats can handle.

My question is would I ever get "Infinity" as the output if Java computes a value that is bigger than what the data type in use can handle? Or is "Infinity" only returned if I have accidentally divided by zero somewhere.

Pang
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Inertial Ignorance
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  • Of course you can get an Infinity from out-of-range values - try multiplying a number by 2 repeatedly, for example. – jasonharper Apr 18 '17 at 16:42
  • But if you multiply a number by 2 repeatedly, you wouldn't ever reach infinity. You would just get a really large number that the data type of your variable can't handle. – Inertial Ignorance Apr 18 '17 at 16:43
  • Possible duplicate of [What are the INFINITY constants in Java, really?](http://stackoverflow.com/questions/13317566/what-are-the-infinity-constants-in-java-really) – John Coleman Apr 18 '17 at 16:49

3 Answers3

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The JLS answers this pretty explicitly.

A floating-point operation that overflows produces a signed infinity.

You are probably dividing by zero somewhere but you could check.

double b = 1.1*Double.MAX_VALUE;
System.out.println(b);

That shows up as infinity.

matt
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  • Would "overflows" mean it reaches a number that is too big for a float to handle? And then Java outputs "Infinity" as a result? – Inertial Ignorance Apr 18 '17 at 16:46
  • I edited the answer to include an example. Yes, if the number is too big it will become infinity. If it is too big negative it will become negative infinity. – matt Apr 18 '17 at 16:48
  • Great, thanks. So I could be getting "Infinity" either because I'm dividing by zero, or because I'm computing a value too large for a float to handle? – Inertial Ignorance Apr 18 '17 at 16:49
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Please refer to thedayofcondor's answer here: What are the INFINITY constants in Java, really?

When your operations leave the range of float, you DO get infinity as a result, by default. As you said, however, infinity usually begins with a zero division somewhere, or with a cyclic (or too large of a) operation. If you post a snippet of your code i can give you a hint on where it's 'unprotected'. 

   Operation            Result
n ÷ ± Infinity           0
±Infinity × ±Infinity   ±Infinity
±nonzero  ÷ 0           ±Infinity
Infinity  + Infinity    Infinity
±0 ÷ ±0                 NaN
Infinity - Infinity     NaN
±Infinity ÷ ±Infinity   NaN
±Infinity × 0           NaN
Community
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Marcel Oliveira
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Yes, according to this: Maximum value for Float in Java? the max value a float can have is Float.MAX_VALUE.

You might want to do some simple checks with the debugger where ever you're dividing this float value. Alternatively, you could sprinkle your code with log statements.

There's a class called StackTraceElement that's provided by the Thread class. You could use that to see which class/line number.

https://docs.oracle.com/javase/7/docs/api/java/lang/Thread.html#getStackTrace() https://docs.oracle.com/javase/7/docs/api/java/lang/StackTraceElement.html

Say you have something like this:

float result = A / B;

Anytime you see something like this, you could use this simple print statement:

float result = A / B;
System.out.format("Divisor: %d, Dividend: %d, result: %d", A, B, result);
// alternatively, add in the stack trace stuff.

Once you find the culprit, the cause is probably pretty obvious. My thoughts are that you might be initializing this 'B' value to 0, but never setting it to the float you actually intend to divide by.

brandon
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