在Google Play的当机报告中获取Random.nextint的IllegalArgumentException

Question

I was looking at my crash report and it says the following report

java.lang.IllegalArgumentException in java.util.Random.nextInt

I do not understand this, nextint takes only a int of the max random number to return, and java compiler will not allow you to put in a different type.

How could this exception be going off?????

Answer 1

仅当给定的值是非正数时，Nextint才会引发IllegalArgumentException。

Answer 2

If in doubt, you can always grep the JDK. In this instance, you can look up Random.nextInt here . Looking at the source code, IllegalArgumentException is thrown if and only if the given integer is not positive.

For the sake of completeness, the relevant code reads:

public int nextInt(int n) {
   if (n <= 0)
     throw new IllegalArgumentException("n must be positive");

   if ((n & -n) == n)  // i.e., n is a power of 2
     return (int)((n * (long)next(31)) >> 31);

   int bits, val;
   do {
       bits = next(31);
       val = bits % n;
    while (bits - val + (n-1) < 0);
   return val;
 }}

Edit: The OP asked a question in the comments,

could you explain "grep the JDK"?

Here I mean a search on grepcode.com for "Random.nextInt".