What EXACTLY does the length operator do?
Many people have been arguing with me with this kind of code:
print(#{"1", "2", "3"})
The output:
3
Now, people argue that the octothorpe (#) returns the number of values in a table, and it does. They're correct.
The other day, I met somebody who thought they could return the number of values in a DICTIONARY:
print(#["oof"])
He, of course, was wrong and got an error. But he then argued that...
print(#"oof")
...is the same as...
print(string.len("oof"))
Both lines return the number 3. However, I argued that the octothorpe works on both strings and tables, and I was right (well, duh).
The very next thing he argues about is this:
print(#{"a", [3] = "b"})
He thinks that the output is 5, because he claimed that "Since there are 2 strings and 1 number in square brackets, the output is 5." BUT, the output is actually ONE. Why? The octothorpe returns only the last bracket key for tables starting with 1. Because [3] is the only bracket value in the table, the output is 1 and not 5.
So how exactly does this work? I really want this guy to stop annoying me.
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1 answer
Let me clear it up then
1) Tables
When # is used on a table, it returns the highest numerical index it can find without stumbling upon nil.
For example
local a = {"hello", "world"}
print(#a)
will print 2, because a[3] is nil.
Another example
local a = {"hello"}
a[3] = "hi"
print(#a)
will print 1, because a[2] is nil.
Note that the # operator will not take into account any dictionairy values.
2) Strings
Yes, # works exactly the same way string.len does
Also to add to @MCAndRobloxUnited's comment, doing t[#t + 1] = val is not using addition on a table. It can be used as a substitute for table.insert. Let me explain:
local t = {"hello"} --table's length aka #t is 1
t[#t + 1] = "world" --#t + 1 is 2, so this is equal to t[2] = "world", as you can see, it inserts the value at the end of the table
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