80-Bus News


April–June 1982, Volume 1, Issue 2

Page 28 of 55


By having counted down to 0, the loop leaves HL pointing at the start of the word in question. DE is then loaded to point to WBUF, and the characters are copied from the location pointed to by HL into the location pointed to by DE. HL and DE are both incremented to point to the next character in the word table and WBUF resepctively, whilst B is incremented to count the letters. Notice the copy process is so arranged that the 00 word delimiter gets copied into WBUF along with the word. That saves me having to make the deliberate effort to put an 0O at the end of the word in WBUF. By copying the 00 across to WBUF, B now contains a count one count greater than the length of the word. So long as we remember this that’s ok.

Now to the cheeky bit. We want to tell people how many characters there are in the word (remember we’ve got the count plus one in B). Now as the word length cannot exceed 9, dictated by the length of WBUF and hopefully no-one has added any words longer than nine characters, and as I suspect that the program might get upset if we set up a word of no length, consisting of 00 only, the word length cannot be less than 1, this means we have a figure between 02H and 0AH in B. If we add 2FH to this, we end up with the ASCIT character for the numbers 1 to 9. So, what about stuffing this value into the text string that is to display it. Well what about it? Now I’m sure I’ve mentioned (and if not, I’m about to) that writing code that alters itself is bad practice, if only for the reason that you can’t put it into EPROM (there are other reasons, but they’re a bit deep). So how do I defend myself for suggesting naughty practices to you. Simple I don’t, I say tut – tut, shake my head and look the other way. Who the hell wants Hangman in EPROM anyway. So the next bit does just that and displays the message.

Here starteth the main loop. This is where we test to see if the trys equals the maximum set and send a ‘lose’ message, otherwise the program goes on to get an input character, decide if it’s been used before, and if not, to compare and fill in the blanks in the word. Sounds simple, and it really is.

Ok, so compare the number of trys taken (from TRYS) with NTRYS. If it’s not there yet, the program skips past the ‘lose’ bit and carries on. Have you noticed my use of uninspired labels for the main program flow. Simply CONT and LOOP. Sorry, but having. said labels should be meaningful, there’s not much else you can do in a linear flow program. Anyway, if the number of trys equals NTRYS, display the ‘sorry you lose’ message. Now here we have a label RSTART, which is where the program restarts in the event of a win. Note that the program goes through this bit, win or lose. This displays the word, but we can’t simply point HL at it and call SNDTXT as it will have gome or all bit 7’s set, which will display funny graphics characters. Instead, LOOP4 gets the characters one by one (up to an 00, as before) these are ANDed with 7FH which is a bit mask to strip the flag bit (see last episode) and then sends them to OUTCHR one by one. .

In the event of a win or a lose, the next thing is to discover if the game is to continue. The appropriate message is displayed and the input routine, GETCHR, is called. When a character is returned (caused by you hitting a key), the character is compared with ‘Y’, if it is that character and only that character (no mucking about with lower case here) then the program goes right back to the beginning and starts all over (no finesse about that bit either). Any character other than ‘y’ will cause the program to jump to the EXIT routine and finish the game.

If the number of trys doesn’t equal NTRYS, then we must display a message inviting an input character, and then call GETCHR to get an input. Having got the input character, the character is compared with 03H (which just happens to be control/C, the CP/M escape character. For the NAS version the byte at OE55H could be changed to 1BH to change the tested into and ‘ESCape’ character to be consistent with other ‘NAS tyye’ programs.). If the character is 03H (or whatever) then the program jumps to the EXIT routine to finish the game. Otherwise the character is taken as valid, so a newline must be sent. Now the CRLF1 subroutine will corrupt both the AF

This is an OCR’d version of the scanned page and likely contains recognition errors.

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