CODEWITHSUNDEEP
lua
solikang
solikang

Reputation: 53

wrong lua table size when using string key index

Usually for getting table size, the standard table library function # operator works.

However when I make a table which has a string key index, it doesn't work.

local function addWriterIdListToTable()

        local returnTable = {}
        local requestString = "1234:16 5678:8 9012:1"
        local idList = requestString:split(" ")
        for i,v in ipairs(idList) do
                local oneId = v:split(":")
                returnTable[oneId[1]] = oneId[2]
        end

        for k,v in pairs(returnTable) do
                print (k .. " " .. v)
        end
        print("size of table: " .. #returnTable)

        return returnTable

end

I want to trsnform a string to table. The function "split" parse a string, split it with parameter as a delimiter, and return as table.

The result of a excution above function like below.

1234 16
9012 1
5678 8
size of table: 0

It shows the content of table exactly as I expected, but its count is not.

Anybody to help me?

Thanks in advance.

Upvotes: 2

Views: 2364

Answers (1)

Lily Ballard
Lily Ballard

Reputation: 185671

The # operator tells you the highest numeric index in the table. If there are any gaps in the numeric indexing, it may return the highest below the gap. Basically, the # operator only works right if you're treating your table like a dense array.

If you actually want to know how many entries are in a table, you'll need to iterate over it using the pairs() function and count how many items you get.

function countTableSize(table)
    local n = 0
    for k, v in pairs(table) do
        n = n + 1
    end
    return n
end

Although I do wonder why you even need to know how many entries are in the table. Typically all you care about is if the table is empty or not, and you can check that by just seeing if next(table) == nil.

Upvotes: 9

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