A very quick question about Lua and variable reassignment

[dizzykiwi3]

This may be a very simple question, but I'm just curious

say I have a function setit

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function setit(x)
x = 5 end

if I define a variable bob to be 10 and pass it through setit, bob will still be equal to 10. Why is this the case?

[markgo]

It's basically the same as this

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bob = 10
function setit()
  local x = bob
  x = 5
end

[HugoBDesigner]

The problem is that, the way you're doing it, you can't detect which variable you're trying to change. For example:

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variable1 = 12
variable2 = "something"

function setit(x)
x = 10
end

When you call it with the variable "variable1", it's gonna send only the variable value, not it's name, so it'd be like:

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setit(variable1)

function setit(local x = 12) --notice how it sets a new variable instead of detecting the one used
x = 10 --now the last "x" (which is a local variable) is the only thing changed
end --After you end the function, the local variable is cleaned, so no change whatsoever has been made in any variable

A nice thing about Lua, though, is that setting table values DOES change the original value, so for instance:

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variable1 = {["a"] = 12, ["b"] = 5}
function setit(x)
x = 10
end

setit(variable1.a) --variable1.a is now 10

Luckily for us, variables are stored in a table called _G, so you can still change a variable's specific value, but you'll have to send the variable's name as a string instead:

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variable1 = 20

function setit(x)
_G[x] = 10
end

setit("variable1") --Now variable1 is 10!

I hope I could make this clear, and I hope I helped too

[dizzykiwi3]

That was outrageously clear and concise Hugo, thanks a bunch!

[dizzykiwi3]

The code for the table values though doesn't seem to be working for me

Code: Select all

variable1 = {["a"] = 12, ["b"] = 5}
function setit(x)
x = 10
end

setit(variable1.a) --variable1.a is now 10

when I copy and paste that into my local console, variable1.a still returns 12

[HugoBDesigner]

Glad I could help
One thing to add, though: This function can be obsolete in most cases, since you can simply change the variable you want to the new value:

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variable1 = 12

--later in the code

variable1 = 10

But this is mostly useful if you need to change multiple variables at once:

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local toChange = {"variable1", "variable2", "variable3", "variable4", ... }
for i, v in ipairs(toChange) do
setit(v)
end

Other than this case, I don't see many needs for that. Since the function has only a single line of code, you could replace any call of it with it's code too:

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local toChange = {"variable1", "variable2", "variable3", "variable4", ... }
for i, v in ipairs(toChange) do
_G[v] = 10
end

But that doesn't mean that what you're trying to do won't work. Just use whatever you feel comfortable with, I just thought I could let this post here, in case you feel like you could use something else


EDIT:

The code for the table values though doesn't seem to be working for me

Code: Select all

variable1 = {["a"] = 12, ["b"] = 5}
function setit(x)
x = 10
end

setit(variable1.a) --variable1.a is now 10

when I copy and paste that into my local console, variable1.a still returns 12

Oh, I'm sorry. Forgot that you actually need to specify the table. So this:

Code: Select all

variable1 = {["a"] = 12, ["b"] = 5}
function setit(x)
for i, v in pairs(x) do
v = 10
end
end

setit(variable1)

Is what would change the values. In this specific example, it changes all the values in the table, but you could also make it change only a single value, or reassemble the entire table, if you REALLY need to:

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variable1 = {12} --variable1[1] is 12

function setit(x)
x = {10}
end

setit(variable1) --variable1[1] is now 10

[Robin]

Oh, I'm sorry. Forgot that you actually need to specify the table. So this:

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variable1 = {["a"] = 12, ["b"] = 5}
function setit(x)
for i, v in pairs(x) do
v = 10
end
end

setit(variable1)

Is what would change the values. In this specific example, it changes all the values in the table, but you could also make it change only a single value, or reassemble the entire table, if you REALLY need to:

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variable1 = {12} --variable1[1] is 12

function setit(x)
x = {10}
end

setit(variable1) --variable1[1] is now 10

Wrong and wrong. (I'm sorry, you were being so helpful to dizzykiwi3)

In Lua, = is "bind", not "assign". If you do avar = anexpression, the only thing that changes is where avar points to. No table is modified, even if it previously pointed to such a table (or a value pointed to by some table).

If a.b is a table, and you do a.b = {}, then only the table a is changed. The Table Previously Known As a.b is unaffected.

You can try it out:

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a = {b = {10}}
local original_table = a.b
print(a.b[1]) -- 10
a.b = {5)
print(a.b[1]) -- 5
print(original_table[1]) -- 10

[HugoBDesigner]

Welp, it turns out I shouldn't have left the Lua school early
Seriously, though, I thought I had these table-persistence things understood, but I guess not. Thanks for the correction, though. I'll read a bit more about Lua tables and variables, just so I get these things properly understood

[dizzykiwi3]

So... what should I do to accomplish what I need?

[HugoBDesigner]

This will still work:

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variable1 = 20

function setit(x)
_G[x] = 10
end

setit("variable1") --Now variable1 is 10!