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((join [1.] [2])- (1))

Outputs 0.199999999999996 (not sure how many 9's).

Considering this is not possible in most other languages the fact that this even compiles is cause for celebration (this is one of the best things about Scratch imo).

Even so, it would be nice if it equaled 0.2. Would this be possible with a quick patch, or is the problem a bit more complicated than that (guessing the latter, but figured I'd ask anyways)?

Yours truly,
-mCC

1/3 = 0.33333333 recurring be represented exactly in base 10.
So if we had 8 decimal places 2 x 1/3 = 0.66666667

Computers use base 2 and cannot represent any decimal if, as a fraction, its denominator is not a factor of 2.
So 1/2, 1/4, 3/8, 7/16 etc can be represented exactly in binary.

But 0.2 = 1/5 cannot. It has an infinite “binimal” part (like a decimal part but base 2).
So 1.2 cannot be represented exactly in base 2.

If you want to correct any number to 1 decimal place:

set [Number v] to ((round ((Number) * (10))) / (10))

eg. 0.199999999999996 * 10 = 1.99999999999996
round(1.99999999999996) = 2
2/10 = 0.02

Scratch-Minion wrote:

1/3 = 0.33333333 recurring be represented exactly in base 10.
So if we had 8 decimal places 2 x 1/3 = 0.66666667

Computers use base 2 and cannot represent any decimal if, as a fraction, its denominator is not a factor of 2.
So 1/2, 1/4, 3/8, 7/16 etc can be represented exactly in binary.

But 0.2 = 1/5 cannot. It has an infinite “binimal” part (like a decimal part but base 2).
So 1.2 cannot be represented exactly in base 2.

If you want to correct any number to 1 decimal place:

set [Number v] to ((round ((Number) * (10))) / (10))

eg. 0.199999999999996 * 10 = 1.99999999999996
round(1.99999999999996) = 2
2/10 = 0.02

Ok, that's a really good workaround. I'll keep that in mind for future projects, thanks!