[FIXED] Why is printing "B" dramatically slower than printing "#"?

Issue

I generated two matrices of 1000 x 1000:

First Matrix: O and #.
Second Matrix: O and B.

Using the following code, the first matrix took 8.52 seconds to complete:

Random r = new Random();
for (int i = 0; i < 1000; i++) {
    for (int j = 0; j < 1000; j++) {
        if(r.nextInt(4) == 0) {
            System.out.print("O");
        } else {
            System.out.print("#");
        }
    }

   System.out.println("");
 }

With this code, the second matrix took 259.152 seconds to complete:

Random r = new Random();
for (int i = 0; i < 1000; i++) {
    for (int j = 0; j < 1000; j++) {
        if(r.nextInt(4) == 0) {
            System.out.print("O");
        } else {
            System.out.print("B"); //only line changed
        }
    }

    System.out.println("");
}

What is the reason behind the dramatically different run times?

---

As suggested in the comments, printing only System.out.print("#"); takes 7.8871 seconds, whereas System.out.print("B"); gives still printing....

As others who pointed out that it works for them normally, I tried Ideone.com for instance, and both pieces of code execute at the same speed.

Test Conditions:

• I ran this test from Netbeans 7.2, with the output into its console
• I used System.nanoTime() for measurements

Solution

Pure speculation is that you're using a terminal that attempts to do word-wrapping rather than character-wrapping, and treats B as a word character but # as a non-word character. So when it reaches the end of a line and searches for a place to break the line, it sees a # almost immediately and happily breaks there; whereas with the B, it has to keep searching for longer, and may have more text to wrap (which may be expensive on some terminals, e.g., outputting backspaces, then outputting spaces to overwrite the letters being wrapped).

But that's pure speculation.

Answered By - T.J. Crowder
Answer Checked By - Gilberto Lyons (JavaFixing Admin)