如何限制（iterator.hasNext（））迭代次数？

Question

I am working on Java using the Generex library, to print strings against given regular expressions .

Some of the R.Es can generate infinite strings , I just want to handle them, but couldn't yet. My code looks like;

Generex generex = new Generex(regex);
Iterator iterator = generex.iterator();
    System.out.println("Possible strings against the given Regular Expression;\n");
    while (iterator.hasNext()) {
        System.out.print(iterator.next() + " ");
    }

If I input (a)* as a regular expression, the output should look like this

a aa aaa aaaa aaaaa aaaaaa aaaaaaa aaaaaaaa aaaaaaaaa ...

How do I limit the result of that loop?

Answer 1

Let's say you wish to print the first 8 items, and then add "..." if there are more items to print. You can do it as follows:

int limit = 8;
int current = 0;
while (iterator.hasNext()) {
    if (current != 0) {
        System.out.print(" ");
    }
    System.out.print(iterator.next());
    // If we reach the limit on the number of items that we print,
    // break out of the loop:
    if (++current == limit) {
        break;
    }
}
// When we exit the loop on break, iterator has more items to offer.
// In this case we should print an additional "..." at the end
if (iterator.hasNext()) {
    System.out.print(" ...");
}

Answer 2

In your case, I think the length of the string is much more important than the number of elements printed so I would say the following solution is probably better :

Generex generex = new Generex(regex);
Iterator iterator = generex.iterator();
System.out.println("Possible strings against the given Regular Expression;\n");
StringBuilder sb = new StringBuilder();
int limitOfChars = 100; //for example
while (iterator.hasNext()) {
    String next = iterator.next();
    if (sb.length() + next.length() > limitOfChars) break;
    sb.append(next + " ");
}
System.out.println(sb.toString() + " ... ");