In C++ you can apply the prefix “static” to variables such that every time the variable is redefined, the system does not attempt to allocate any more memory to it.
I assume something similar would be the case in Luau, since creating new instances is far slower than reusing existing ones.
Running the following code:
local t = tick()
local a
for i = 0, 1e9 do
a = i
end
print(tick() - t) -- Time taken in seconds: 6.323152780532837
local t = tick()
for i = 0, 1e9 do
local a
a = i
end
print(tick() - t) -- Time taken in seconds: 2.804739236831665
Contrary to my expectations, reusing variables is over 2 times slower than defining new ones.
I have yet to check the memory consumption since I’m not too sure how to go about doing that.
Feel free to replicate this yourself and share your result.
This does not appear to be the case when I benchmark it. In fact, declaring a and then assigning it within the same iteration is actually the slowest, while directly assigning a is the fastest.
The result of declaring a outside of the loop and then constantly redefining it falls within the margin of error as well.
["local a; loop; a = i"] = function(Profiler)
local a
for i = 1, 1e7 do
a = i
end
end;
["local a; a = i"] = function(Profiler)
for i = 1, 1e7 do
local a
a = i
end
end;
["local a = i"] = function(Profiler)
for i = 1, 1e7 do
local a = i
end
end;
This is partially because of the way Lua’s scopes work. The first example requires the scope to obtain a non-constant upvalue (a), alter it and push an update back to it.
The second one on the other hand, creates a localized variable that gets changed once and GC’d after the scope ends. It doesn’t need to push any update between scopes because a will only exist in that scope.
Do note these results are entirely deterministic and you shouldn’t try to incorporate micro-optimizations like this, since chances are as time goes on the better Luau’s VM will handle these.
