[ANN] Lua 5.5.0 now available

Luiz Henrique de Figueiredo

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Thank you and the whole team for this christmas present!

- mb

Lua 5.5.0 has been frozen and is now the current version of Lua.

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> On 22 Dec 2025, at 15:10, Luiz Henrique de Figueiredo <l...@tecgraf.puc-rio.br> wrote:
>
> Lua 5.5.0 has been frozen and is now the current version of Lua.

Thank you and congratulations. Happy holidays!

~b

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Jason Lethbridge

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Is " large arrays use about 60% less memory" right?

Lua 5.4 and earlier stored arrays as 8 byte value, 1 byte type

tag, and 7 byte padding, so 16 bytes per value.

Lua 5.5 splits those and packs them tightly, so the 7 byte padding

is gone, so it's just 8 + 1 = 9 bytes per value.

This means 5.5 has 43% (7 div 16) smaller arrays, so

they take 57% of space they did before, not 60% less.

And thank you for this release - B.R.

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Thanks all who develop and/or use Lua!  Congratulations! ^.^

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Congratulations with release and merry Christmas!

-- Martin

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Hi Lua Team,

In the part of the manual related to
changes ( https://www.lua.org/manual/5.5/readme.html#changes ),
it says:

> for-loop variables are read only

I performed a few tests on that
matter with Lua 5.5 on
Linux + Windows, and I have a doubt.

1. Executing the following statement:

    lua -e "for i, v in ipairs({1}) do i = 3 end"

    it throws the error
    `lua: (command line):1: attempt to assign to const variable 'i'`

2. Executing this statement:

    lua -e "for i, v in ipairs({1}) do v = 3 end"

    it does not throw any error.
    Then I thought: hmm, probably
    it assigns the value back.
    However, the next test shows
    it does not.

3. Executing this statement:

    lua -e "local t = {1}; for i, v in ipairs(t) do v = 3 end; print(t[1]);"

    prints 1, showing that the
    t[1] value didn't change.

So, my question is: according to
the manual

> for-loop variables are read only

the second part above (2.) shouldn't throw?

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Luau Project <luau.p...@gmail.com> 于2025年12月26日周五 07:54写道：
>
> Hi Lua Team,

The manual says "The first of these variables is the control variable,
which is a read-only (const) variable."

So the "i" is ready-only (control variable) , but the "v" is not.

--
http://blog.codingnow.com

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Damn, I just wanted to paste same quote from refman. You're ~2 minutes
faster

[[

A *for* statement like

for/var_1/, ···,/var_n/ in/explist/ do/body/ end

works as follows.

The names /var_i/ declare loop variables local to the loop body. The
first of these variables is the /control variable/, which is a read-only
(|const|) variable.

]]

-- Martin

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Thank you all for the insightful answers.

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That's because v is a variable holding a value, not a reference.

Similarly:

t = { 1 }
a = t[1]
a = 2
print(t[1], a)
-- 1 2

Scott