BASICally, Its Retro Machine Language
We enjoyed [Beej’s] trip down memory lane looking at a BASIC game, The Wizard’s Castle, written for the Exidy Sorcerer. It appeared in a 1980 magazine that included the title …read more

We enjoyed [Beej’s] trip down memory lane looking at a BASIC game, The Wizard’s Castle, written for the Exidy Sorcerer. It appeared in a 1980 magazine that included the title graphic above. It reminded us how, back in those days, we did things with BASIC that you shouldn’t be able to do and it often looks, today, rather cryptic.
In particular, even if you know modern BASIC, these few lines might give you a pause: 10 REM"_(C2SLFF4 40 POKE 260,218: POKE 261,1: T = USR(0): T = PEEK(-2049) 80 Q = RND(-(2*T+1)) Line 10 is a comment, but a strange one. Certainly that doesn’t matter, right? Actually, it is a key part of the action.
On line 40, you can see some pokes to write directly to memory and a peek to read some memory value back. The USR function calls some machine language program. You may realize the whole thing is to get some value T to seed the random number generator in line 80.
This leads to a few obvious questions. First, how does USR know what to call? Second, where is the machine language program?
The details varied by system, of course, but in this case, the program knows that location 259 has a jump instruction that USR called. So poking an address into 260 and 261 was telling USR where it should go. But what’s at that address?
Keep in mind that an old computer like the Sorcerer didn’t have megabytes of memory being swapped about by an operating system. That means that things tended to be in known places and that BASIC had to be judicious about storing source code. As was common at the time, a line like “10 PRINT 1+1” would get tokenized.
In this case, each line would get a pointer to the next line, a two-byte line number, a single-byte token for “PRINT” and then more bytes to represent the rest of the line. In the case of text in a string or a remark, the bytes were just the text with a zero to terminate the string. The first line entered would always be at address 469.
So? If you consider the format of the REM statement, there will be a pointer at 469 an
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