假设有一个方法用于通过传递url来获取字符串中的数据：
public static void main(String... args) {
Random r = new Random();
Iterator<Long> it = randomSequence(r, 10, 32);
while(it.hasNext()) {
System.out.println(it.next());
}
}
/**
* Random sequence generator.
*
* @param r the random generator
* @param size the number of entries to generate
* @param shift the number of bits of the result
* @return the iterator
*/
static Iterator<Long> randomSequence(final Random r, final long size, final int shift) {
if (size < 5) {
// small lists are generated using a regular hash set
TreeSet<Long> set = new TreeSet<Long>();
while (set.size() < size) {
set.add(r.nextLong() & ((2L << shift) - 1));
}
return set.iterator();
}
// large lists are created recursively
return new Iterator<Long>() {
long remaining = size, zeros = randomHalf(r, size);
Iterator<Long> lowBits0 = randomSequence(r, zeros, shift - 1);
Iterator<Long> lowBits1;
@Override
public boolean hasNext() {
return remaining > 0;
}
@Override
public Long next() {
remaining--;
if (lowBits0.hasNext()) {
return lowBits0.next();
}
if (lowBits1 == null) {
lowBits1 = randomSequence(r, size - zeros, shift - 1);
}
return (1L << shift) + lowBits1.next();
}
};
}
/**
* Get the number of entries that are supposed to be below the half,
* according to the probability theory. For example, for a number of coin
* flips, how many are heads.
*
* @param r the random generator
* @param samples the total number of entries
* @return the number of entries that should be used for one half
*/
static long randomHalf(Random r, long samples) {
long low = 0, high = samples;
double x = r.nextDouble();
while (low + 1 < high) {
long mid = (low + high) / 2;
double p = probabilityBucketAtMost(samples, mid);
if (x > p) {
low = mid;
} else {
high = mid;
}
}
return (low + high) / 2;
}
static double probabilityBucketAtMost(long flips, long heads) {
// https://www.fourmilab.ch/rpkp/experiments/statistics.html
long x = heads;
long n = flips;
double variance = Math.sqrt(n/4);
// mean
long mu = n / 2;
// https://en.wikipedia.org/wiki/Normal_distribution
// Numerical approximations for the normal CDF
// the probability that the value of a standard normal random variable X is <= x
return phi((x - mu) / variance);
}
static double phi(double x) {
return 0.5 * (1 + Math.signum(x) * Math.sqrt(1 - Math.exp(-2 * x * x / Math.PI)));
}
现在，如果我从三到四个AsyncTasks调用此方法并传递不同的URL，是否有可能导致数据不匹配..