Sunday, November 29, 2009 12:15 AM,
A student writes, although the problem he identifies is probably not a problem for most of you. If you have no trouble seeing ten instances of the character you typed interrupting the UT checkerboard display, like that shown on the screen shot, feel free to delete this message and move on. Good Evening Dr. Patt, I believe I have completed programming assignment 5, but I'm finding it hard to see whether or not the typed key was really displayed 10 times. When the program is running and I press a key I do see that the console changes and I'm able to get a quick glance of the key that was typed but by the time I stop execution the repeated string is gone and so I'm not really confident of what took place during the program. I tried augmenting the delay (changing .FILL #2500 to .FILL #65000) but the output at the console is still being refreshed way too fast. Perhaps I'm not implementing the delay subroutine correctly? Thank you. <<name withheld to protect the student with a superfast machine, maybe>> So far, only two students have had this problem, and the problem went away by adding delay to the DELAY subroutine you are to call after each line is output to the screen. We suggested .FILL #2500 since that should be enough to slow down the display of the checkerboard, so that the interrupted character doesn't rise up and off the screen before you can count how many characters the interrupt routine displays. If not, try .FILL xF000. If still not enough, put a JSR DELAY2 inside the loop of DELAY. That will cause the loop of DELAY to count once for every n iterations of the loop inside DELAY2, where DELAY2 contains .FILL n. So, for example, if DELAY has .FILL #1200 and DELAY2 has .FILL #1300, the delay would be 1,560,000, and that should be enough. If not, try bigger values. That is, if you happen to have a superfast laptop, I did not think it would be the case, but I am told the UT checkerboard comes out much too fast. Hope this helps. Good luck getting prog5 done in time. Yale Patt