I'm writing a LSL to Lua translator, and I'm having all sorts of trouble implementing incrementing and decrementing operators. LSL has such things using the usual C like syntax (x++, x--, ++x, --x), but Lua does not. Just to avoid massive amounts of typing, I refer to these sorts of operators as "crements". In the below code, I'll use "..." to represent other parts of the expression.
... x += 1 ...
Wont work, coz Lua only has simple assignment.
... x = x + 1 ...
Wont work coz that's a statement, and Lua can't use statements in expressions. LSL can use crements in expressions.
function preIncrement(x) x = x + 1; return x; end
... preIncrement(x) ...
While it does provide the correct value in the expression, Lua is pass by value for numbers, so the original variable is not changed. If I could get this to actually change the variable, then all is good. Messing with the environment might not be such a good idea, dunno what scope x is. I think I'll investigate that next. The translator could output scope details.
Assuming the above function exists -
... x = preIncrement(x) ...
Wont work for the "it's a statement" reason.
Other solutions start to get really messy.
x = preIncrement(x)
... x ...
Works fine, except when the original LSL code is something like this -
while (doOneThing(x++))
{
doOtherThing(x);
}
Which becomes a whole can of worms. Using tables in the function -
function preIncrement(x) x[1] = x[1] + 1; return x[1]; end
temp = {x}
... preincrement(temp) ...
x = temp[1]
Is even messier, and has the same problems.
Starting to look like I might have to actually analyse the surrounding code instead of just doing simple translations to sort out what the correct way to implement any given crement will be. Anybody got any simple ideas?
I think to really do this properly you're going to have to do some more detailed analysis, and splitting of some expressions into multiple statements, although many can probably be translated pretty straight-forwardly.
Note that at least in C, you can delay post-increments/decrements to the next "sequence point", and put pre-increments/decrements before the previous sequence point; sequence points are only located in a few places: between statements, at "short-circuit operators" (&& and ||), etc. (more info here)
So it's fine to replace x = *y++ + z * f (); with { x = *y + z * f(); y = y + 1; }—the user isn't allowed to assume that y will be incremented before anything else in the statement, only that the value used in *y will be y before it's incremented. Similarly, x = *--y + z * f(); can be replaced with { y = y - 1; x = *y + z * f (); }
Lua is designed to be pretty much impervious to implementations of this sort of thing. It may be done as kind of a compiler/interpreter issue, since the interpreter can know that variables only change when a statement is executed.
There's no way to implement this kind of thing in Lua. Not in the general case. You could do it for global variables by passing a string to the increment function. But obviously it wouldn't work for locals, or for variables that are in a table that is itself global.
Lua doesn't want you to do it; it's best to find a way to work within the restriction. And that means code analysis.
Your proposed solution only will work when your Lua variables are all global. Unless this is something LSL also does, you will get trouble translating LSL programs that use variables called the same way in different places.
Lua is only able of modifying one lvalue per statement - tables being passed to functions are the only exception to this rule. You could use a local table to store all locals, and that would help you out with the pre-...-crements; they can be evaluated before the expression they are contained in is evauated. But the post-...-crements have to be evaluated later on, which is simply not possible in lua - at least not without some ugly code involving anonymous functions.
So you have one options: you must accept that some LSL statements will get translated to several Lua statements.
Say you have a LSL statement with increments like this:
f(integer x) {
integer y = x + x++;
return (y + ++y)
}
You can translate this to a Lua statement like this:
function f(x) {
local post_incremented_x = x + 1 -- extra statement 1 for post increment
local y = x + post_incremented_x
x = post_incremented_x -- extra statement 2 for post increment
local pre_incremented_y = y + 1
return y + pre_incremented_y
y = pre_incremented_y -- this line will never be executed
}
So you basically will have to add two statements per ..-crement used in your statements. For complex structures, that will mean calculating the order in which the expressions are evaluated.
For what is worth, I like with having post decrements and predecrements as individual statements in languages. But I consider it a flaw of the language when they can also be used as expressions. The syntactic sugar quickly becomes semantic diabetes.
After some research and thinking I've come up with an idea that might work.
For globals -
function preIncrement(x)
_G[x] = _G[x] + 1
return _G[x]
end
... preIncrement("x") ...
For locals and function parameters (which are locals to) I know at the time I'm parsing the crement that it is local, I can store four flags to tell me which of the four crements is being used in the variables AST structure. Then when it comes time to output the variables definition, I can output something like this -
local x;
function preIncrement_x() x = x + 1; return x; end
function postDecrement_x() local y = x; x = x - 1; return y; end
... preIncrement_x() ...
In most of your assessment of configurability to the code. You are trying to hard pass data types from one into another. And call it a 'translator'. And in all of this you miss regex and other pattern match capacities. Which are far more present in LUA than LSL. And since the LSL code is being passed into LUA. Try using them, along with other functions. Which would define the work more as a translator, than a hard pass.
Yes I know this was asked a while ago. Though, for other viewers of this topic. Never forget the environments you are working in. EVER. Use what they give you to the best ability you can.
Related
I'm trying to run the following in Lua 5.3
function new_t()
local self = {}
setmetatable(self, {
__add = function(lhs,rhs)
print('ok so',lhs,'+',rhs)
end
})
return self
end
local t1 = new_t()
local t2 = new_t()
t1 + t2
I get an error saying syntax error near '+'. However if I change the last line to x = t1 + t2, it runs and prints without error.
Is it possible to use a binary operator without using the resulting value? Why doesn't Lua let me do t1 + t2 or even 1 + 2 by itself?
Lua doesn't allow this, because all the operators (except function calls) are intended to always calculate a result. There's no good reason to throw away the result of an expression, and it usually indicates a coding mistake.
If you just want to test your code, I suggest using assert:
assert(not (t1 + t2))
I use not here, because your __add function doesn't return anything.
EDIT: Normally, when we add two numbers, we expect to get a new number, without changing the original numbers. Lua's metamethods are designed to work the same way. To do side-effects like printing or modifying an operand, it's easier and clearer to use a regular named method.
I'd like the example computation expression and values below to return 6. For some the numbers aren't yielding like I'd expect. What's the step I'm missing to get my result? Thanks!
type AddBuilder() =
let mutable x = 0
member _.Yield i = x <- x + i
member _.Zero() = 0
member _.Return() = x
let add = AddBuilder()
(* Compiler tells me that each of the numbers in add don't do anything
and suggests putting '|> ignore' in front of each *)
let result = add { 1; 2; 3 }
(* Currently the result is 0 *)
printfn "%i should be 6" result
Note: This is just for creating my own computation expression to expand my learning. Seq.sum would be a better approach. I'm open to the idea that this example completely misses the value of computation expressions and is no good for learning.
There is a lot wrong here.
First, let's start with mere mechanics.
In order for the Yield method to be called, the code inside the curly braces must use the yield keyword:
let result = add { yield 1; yield 2; yield 3 }
But now the compiler will complain that you also need a Combine method. See, the semantics of yield is that each of them produces a finished computation, a resulting value. And therefore, if you want to have more than one, you need some way to "glue" them together. This is what the Combine method does.
Since your computation builder doesn't actually produce any results, but instead mutates its internal variable, the ultimate result of the computation should be the value of that internal variable. So that's what Combine needs to return:
member _.Combine(a, b) = x
But now the compiler complains again: you need a Delay method. Delay is not strictly necessary, but it's required in order to mitigate performance pitfalls. When the computation consists of many "parts" (like in the case of multiple yields), it's often the case that some of them should be discarded. In these situation, it would be inefficient to evaluate all of them and then discard some. So the compiler inserts a call to Delay: it receives a function, which, when called, would evaluate a "part" of the computation, and Delay has the opportunity to put this function in some sort of deferred container, so that later Combine can decide which of those containers to discard and which to evaluate.
In your case, however, since the result of the computation doesn't matter (remember: you're not returning any results, you're just mutating the internal variable), Delay can just execute the function it receives to have it produce the side effects (which are - mutating the variable):
member _.Delay(f) = f ()
And now the computation finally compiles, and behold: its result is 6. This result comes from whatever Combine is returning. Try modifying it like this:
member _.Combine(a, b) = "foo"
Now suddenly the result of your computation becomes "foo".
And now, let's move on to semantics.
The above modifications will let your program compile and even produce expected result. However, I think you misunderstood the whole idea of the computation expressions in the first place.
The builder isn't supposed to have any internal state. Instead, its methods are supposed to manipulate complex values of some sort, some methods creating new values, some modifying existing ones. For example, the seq builder1 manipulates sequences. That's the type of values it handles. Different methods create new sequences (Yield) or transform them in some way (e.g. Combine), and the ultimate result is also a sequence.
In your case, it looks like the values that your builder needs to manipulate are numbers. And the ultimate result would also be a number.
So let's look at the methods' semantics.
The Yield method is supposed to create one of those values that you're manipulating. Since your values are numbers, that's what Yield should return:
member _.Yield x = x
The Combine method, as explained above, is supposed to combine two of such values that got created by different parts of the expression. In your case, since you want the ultimate result to be a sum, that's what Combine should do:
member _.Combine(a, b) = a + b
Finally, the Delay method should just execute the provided function. In your case, since your values are numbers, it doesn't make sense to discard any of them:
member _.Delay(f) = f()
And that's it! With these three methods, you can add numbers:
type AddBuilder() =
member _.Yield x = x
member _.Combine(a, b) = a + b
member _.Delay(f) = f ()
let add = AddBuilder()
let result = add { yield 1; yield 2; yield 3 }
I think numbers are not a very good example for learning about computation expressions, because numbers lack the inner structure that computation expressions are supposed to handle. Try instead creating a maybe builder to manipulate Option<'a> values.
Added bonus - there are already implementations you can find online and use for reference.
1 seq is not actually a computation expression. It predates computation expressions and is treated in a special way by the compiler. But good enough for examples and comparisons.
I'm not sure what the point of setting the metamethod of Stack to Stack is.
I've looked around but none of the explanations made too much sense. I get that sometimes instances may need to fall back on Stack for any missing methods but was hoping for a clearer explanation.
Stack = {}
Stack.__index = Stack
Let's say you have a "2d vector" "class" which has a .x and a .y. You write a dotproduct function for them:
function dotproduct(a, b)
return a.x * b.x + a.y * b.y
end
The simplest way to let you write foo:dot(bar) would be to put a .dot property on every new instance:
-- Make a "V2" namespace
local V2 = {}
function V2.new(x, y)
return {
x = x,
y = y,
-- Attach any methods to the object, also
dot = dotproduct,
}
end
However, as the number of methods grows, this gets worse for two reasons. One is that your objects get larger -- even though you only need to hold two numbers on each vector, most of the hashtable is functions which are always the same! Another is that multiple constructors get difficult because you need to duplicate the work in each constructor.
A solution to this is the __index metatable. Instead of copying each method onto each new object, you can just point to where to find "missing" methods as they needed:
V2.dot = dotproduct
function V2.new(x, y)
return setmetatable({x = x, y = y}, {__index = V2})
end
This design has a small drawback. First, you're allocating a new metatable in addition to every table, which is a waste of memory (all of the metatables are the same). Second, you don't have an opportunity to override other methods without growing this new metatable (the same annoyance from before).
So, we just make V2 be the single metatable:
V2.__index = V2
function V2.new(x, y)
return setmetatable({x = x, y = y}, V2)
end
If I have a function that returns multiple values, how can I access those values separately? Something like table[i].
angles = function()
x = function()
local value = 0
return value
end
y = function()
local value = 90
return value
end
z = function()
local value = 180
return value
end
return x(), y(), z()
end
A problem arises here when wanting to use, for example, the x value separately, while keeping it in the function angles.
print(????)
Sort of wish functions worked like tables in this respect, so I could type something like print(angles.x)
Also, I know that code seems really redundant, but it's actually a much more simplified version of what I'm actually using. Sorry if it makes less sense that way.
x, y, z= angles()
print (x,y,z)
There's a couple of ways to do this.
Most obvious would be
local x, y, z = angles()
print(x)
If you want the first value specifically
local x = ( angles() )
-- `local x = angles()` would work too. Lua discards excess return values.
print(x)
or, somewhat less readably
print((angles()))
You could also return a table from the function, or use the standard module table to pack the return values into one.
local vals = table.pack(angles())
print(vals[1])
Another way of accessing them individually (as the question wording implies) rather than all at once is this:
print((select(1,angles())))
print((select(2,angles())))
print((select(3,angles())))
Output:
0
90
180
The select() call needs to be in parentheses in order to return individual entries rather than all after the given offset.
I need a base converter function for Lua. I need to convert from base 10 to base 2,3,4,5,6,7,8,9,10,11...36 how can i to this?
In the string to number direction, the function tonumber() takes an optional second argument that specifies the base to use, which may range from 2 to 36 with the obvious meaning for digits in bases greater than 10.
In the number to string direction, this can be done slightly more efficiently than Nikolaus's answer by something like this:
local floor,insert = math.floor, table.insert
function basen(n,b)
n = floor(n)
if not b or b == 10 then return tostring(n) end
local digits = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ"
local t = {}
local sign = ""
if n < 0 then
sign = "-"
n = -n
end
repeat
local d = (n % b) + 1
n = floor(n / b)
insert(t, 1, digits:sub(d,d))
until n == 0
return sign .. table.concat(t,"")
end
This creates fewer garbage strings to collect by using table.concat() instead of repeated calls to the string concatenation operator ... Although it makes little practical difference for strings this small, this idiom should be learned because otherwise building a buffer in a loop with the concatenation operator will actually tend to O(n2) performance while table.concat() has been designed to do substantially better.
There is an unanswered question as to whether it is more efficient to push the digits on a stack in the table t with calls to table.insert(t,1,digit), or to append them to the end with t[#t+1]=digit, followed by a call to string.reverse() to put the digits in the right order. I'll leave the benchmarking to the student. Note that although the code I pasted here does run and appears to get correct answers, there may other opportunities to tune it further.
For example, the common case of base 10 is culled off and handled with the built in tostring() function. But similar culls can be done for bases 8 and 16 which have conversion specifiers for string.format() ("%o" and "%x", respectively).
Also, neither Nikolaus's solution nor mine handle non-integers particularly well. I emphasize that here by forcing the value n to an integer with math.floor() at the beginning.
Correctly converting a general floating point value to any base (even base 10) is fraught with subtleties, which I leave as an exercise to the reader.
you can use a loop to convert an integer into a string containting the required base. for bases below 10 use the following code, if you need a base larger than that you need to add a line that mapps the result of x % base to a character (usign an array for example)
x = 1234
r = ""
base = 8
while x > 0 do
r = "" .. (x % base ) .. r
x = math.floor(x / base)
end
print( r );