CODEWITHSUNDEEP
lualua-table
Деян Добромиров
Деян Добромиров

Reputation: 902

How to check if a value is a table

So simply as the name suggests... You can't say like

3[7] --> Attempt to index a number value
"fpp"[7] --> Attempt to index a string value


I know the function type(), but I am trying to avoid it, because it's slow.

if(type({}) == "table") ...
if(string.sub(tostring({}),1,5) == "table")...

function ArrayCount(arArr)
  if(not arArr) then return 0 end
  if(not (type(arArr) == "table")) then return 0 end
  if(not (arArr and arArr[1])) then return return 0 end
  local Count = 1
  while(arArr[Count]) do Count = Count + 1 end
  return (Count - 1)
end

ArrayCount(3)
ArrayCount("I am a string!")

Upvotes: 1

Views: 2538

Answers (1)

hjpotter92
hjpotter92

Reputation: 80657

The type function is not slow. Update your function to simply:

function ArrayCount(arArr)
  if type( arArr ) ~= "table" then return 0 end
  return #arArr - 1
end

Upvotes: 4

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