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Editorial Team
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Asked: 2026-05-16T16:53:38+00:00

The following two (functionally equivalent) programs are taken from an old issue of Compute’s

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The following two (functionally equivalent) programs are taken from an old issue of Compute’s Gazette. The primary difference is that program 1 puts the target base memory locations (7680 and 38400) in-line, whereas program 2 assigns them to a variable first.

Program 1 runs about 50% slower than program 2. Why? I would think that the extra variable retrieval would add time, not subtract it!

10 PRINT"[CLR]":A=0:TI$="000000"
20 POKE 7680+A,81:POKE 38400+A,6:IF A=505 THEN GOTO 40
30 A=A+1:GOTO 20
40 PRINT TI/60:END

Program 1

10 PRINT "[CLR]":A=0:B=7600:C=38400:TI$="000000"
20 POKE B+A,81:POKE C+A,6:IF A=505 THEN GOTO 40
30 A=A+1:GOTO 20
40 PRINT TI/60:END

Program 2

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1 Answer

  1. Editorial Team

    The reason is that BASIC is fully interpreted here, so the strings “7680” and “38400” need to be converted to binary integers EVERY TIME line 20 is reached (506 times in this program). In program 2, they’re converted once and stored in B. So as long as the search-for-and-fetch of B is faster than convert-string-to-binary, program 2 will be faster.

    If you were to use a BASIC compiler (not sure if one exists for VIC-20, but it would be a cool retro-programming project), then the programs would likely be the same speed, or perhaps 1 might be slightly faster, depending on what optimizations the compiler did.

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