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Can someone explain this?
Please make your question title relevant to your question content. It should be a one-sentence summary in question form.
I'm not sure what and ... how
local t = {"1", "1", "1", "1"}
t[2] = nil
t[4] = nil
print(#t) ---> 1
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2
See here; it says "Frequently, in Lua, we assume that an array ends just before its first nil element."
Now, "#t" is not guaranteed to work this way. The Lua Reference Manual states "The length of a table t is defined to be any integer index n such that t[n] is not nil and t[n+1] is nil" (thanks to BlueTaslem for finding it); the full documentation is here. For instance, t={1,1,1} t[2]=nil print(#t) prints out 3 (though you should not rely on this behaviour, as it could just as easily print out 1).
If you need 'nil' values in a table, you have two options:
- Declare
nilValue = {}and use that instead ofnil - Never use
#on the table; usepairsto iterate over the table (notably, this will skip over nil values)
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Another option is to `table.remove(t, 2)` instead -- this doesn't leave a gap BlueTaslem 18071 — 9y
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"The length of a table t is defined to be any integer index n such that t[n] is not nil and t[n+1] is nil" BlueTaslem 18071 — 9y
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In your example #t is EITHER 1 or 3. ("If the array has "holes" (that is, nil values between other non-nil values), then #t can be any of the indices that directly precedes a nil value") [Lua reference 2.5.5] BlueTaslem 18071 — 9y
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Thanks; good points. I've updated my answer. chess123mate 5873 — 9y