Re: [Haskell-cafe] some questions about Template Haskell
TP wrote:
pr :: Name -> ExpQ pr n = [| putStrLn $ (nameBase n) ++ " = " ++ show $(varE n) |]
The example is indeed problematic. Let's consider a simpler one:
foo :: Int -> ExpQ foo n = [|n + 1|]
The function f, when applied to an Int (some bit pattern in a machine register), produces _code_. It helps to think of the code as a text string with the source code. That text string cannot include the binary value that is n. That binary value has to be converted to the numeric text string, and inserted in the code. That conversion is called `lifting' (or quoting). The function foo is accepted because Int is a liftable type, the instance of Lift. And Name isn't. BTW, the value from the heap of the running program inserted into the generated code is called `cross-stage persistent'. The constraint Lift is implicitly generated by TH when it comes across a cross-stage persistent identifier. You can read more about it at http://okmij.org/ftp/ML/MetaOCaml.html#CSP Incidentally, MetaOCaml would've accepted your example, for now. There are good reasons to make the behavior match that of Haskell.
oleg@okmij.org wrote:
pr :: Name -> ExpQ pr n = [| putStrLn $ (nameBase n) ++ " = " ++ show $(varE n) |]
The example is indeed problematic. Let's consider a simpler one:
foo :: Int -> ExpQ foo n = [|n + 1|]
The function f, when applied to an Int (some bit pattern in a machine register), produces _code_. It helps to think of the code as a text string with the source code. That text string cannot include the binary value that is n. That binary value has to be converted to the numeric text string, and inserted in the code. That conversion is called `lifting' (or quoting). The function foo is accepted because Int is a liftable type, the instance of Lift. And Name isn't.
Thanks Oleg, Probably the following question will be stupid, but I ask it anyway: in my initial example, (nameBase n) returns a String, so we are not in the case where it is not "liftable"? In fact I am not sure to have understood your answer. Now, I have found another behavior difficult to understand for me:
runQ $ lift "u" ListE [LitE (CharL 'u') runQ $ [| "u" |] LitE (StringL "u")
So we have similar behaviors for lift and [||]. We can check it in a splice:
$( [| "u" |] ) "u" $( lift "u" ) "u"
But if I replace a working version: pr n = [| putStrLn ( $(lift( nameBase n ++ " = " )) ++ show $(varE n) ) |] by pr n = [| putStrLn ( $([| (nameBase n) ++ " = " |]) ++ show $(varE n) ) |] I again get the error """ No instance for (Lift Name) arising from a use of `n' Possible fix: add an instance declaration for (Lift Name) In the first argument of `nameBase', namely `n' """ It is not easy to surmise the reason for this error. TP
On Mon, Jul 1, 2013 at 6:01 AM, TP
oleg@okmij.org wrote:
pr :: Name -> ExpQ pr n = [| putStrLn $ (nameBase n) ++ " = " ++ show $(varE n) |]
The example is indeed problematic. Let's consider a simpler one:
foo :: Int -> ExpQ foo n = [|n + 1|]
The function f, when applied to an Int (some bit pattern in a machine register), produces _code_. It helps to think of the code as a text string with the source code. That text string cannot include the binary value that is n. That binary value has to be converted to the numeric text string, and inserted in the code. That conversion is called `lifting' (or quoting). The function foo is accepted because Int is a liftable type, the instance of Lift. And Name isn't.
Thanks Oleg, Probably the following question will be stupid, but I ask it anyway: in my initial example, (nameBase n) returns a String, so we are not in the case where it is not "liftable"? In fact I am not sure to have understood your answer.
The problem isn't the output of nameBase, it's the input parameter 'n'. In your example, you've created a function that takes input (a Name) and generates code based upon that input. In order to lift a value (n) from an ordinary context into a quote, it needs a Lift instance.
Now, I have found another behavior difficult to understand for me:
runQ $ lift "u" ListE [LitE (CharL 'u') runQ $ [| "u" |] LitE (StringL "u")
So we have similar behaviors for lift and [||]. We can check it in a splice:
$( [| "u" |] ) "u" $( lift "u" ) "u"
But if I replace a working version:
pr n = [| putStrLn ( $(lift( nameBase n ++ " = " )) ++ show $(varE n) ) |]
by
pr n = [| putStrLn ( $([| (nameBase n) ++ " = " |]) ++ show $(varE n) ) |]
I again get the error
In the working version, 'n' appears inside a splice, whereas in the other n is in a quote. AFAIK any value can be used in a splice (provided it meets the staging restrictions), whereas only Lift-able values can be used in a quote. Perhaps it helps if you think about what a quote does: it allows you to write essentially a string of Haskell code that is converted into an AST. For this to work, the quote parser needs to know how to generate the AST for an identifier. Like much of Haskell, it's type-driven. For identifiers in scope from imports, TH simply generates a variable with the correct name. But for data, the parser needs a way to generate an AST representation, which is what Lift is for.
John Lato wrote:
The problem isn't the output of nameBase, it's the input parameter 'n'. In your example, you've created a function that takes input (a Name) and generates code based upon that input. In order to lift a value (n) from an ordinary context into a quote, it needs a Lift instance.
:i lift class Lift t where
Thanks John. Ok I can understand that a Lift instance is needed, but to use the lift function below, we also need a Lift instance for the return of (nameBase n), because lift is a function that operates on instances of the Lift typeclass: lift :: t -> Q Exp And it is indeed the case:
:i Lift [...] instance Lift a => Lift [a] instance Lift Char
And as I have shown on a small example, lift and [||] return about the same result:
runQ $ lift "u" ListE [LitE (CharL 'u') runQ $ [| "u" |] LitE (StringL "u")
So what is the difference between lift and [||]? Although I feel stupid, I cannot lie and claim I have understood.
Perhaps it helps if you think about what a quote does: it allows you to write essentially a string of Haskell code that is converted into an AST. For this to work, the quote parser needs to know how to generate the AST for an identifier. Like much of Haskell, it's type-driven. For identifiers in scope from imports, TH simply generates a variable with the correct name. But for data, the parser needs a way to generate an AST representation, which is what Lift is for.
Ok, I think I understand that (we need some method to transform a value at data level in a token of an AST), but it seems to me it does not answer my question above. But I am probably wrong. Thanks TP
On Mon, Jul 1, 2013 at 5:42 PM, TP
So what is the difference between lift and [||]? Although I feel stupid, I cannot lie and claim I have understood.
Hi TP, Sometimes [| |] does need to call lift. If for some reason the original lift wasn't exported, you could define: myLift x = [| x |] But if you have enough $( ) splices to balance out the [| |], there is no lift involved: myId1 x = $( [| x |] ) myId2 x = [| $(x) |] I think your first example is supposed to do: *Pr> let x = 5 in $(pr 'x) x = 5 That's possible if you had defined pr as: pr n = [| putStrLn $ $(lift (nameBase n)) ++ " = " ++ show $(varE n) |] If there were no [| |] quotes, but still the ' syntax for getting a Name, it would still be possible to define pr: pr2 n = varE 'putStrLn `appE` (infixE (Just (lift (nameBase n))) (varE '(++)) (Just (infixE (Just (lift " = ")) (varE '(++)) (Just (appE (varE 'show) (varE n)))))) -- Adam
John Lato wrote:
Now, I have found another behavior difficult to understand for me:
runQ $ lift "u" ListE [LitE (CharL 'u') runQ $ [| "u" |] LitE (StringL "u")
So we have similar behaviors for lift and [||]. We can check it in a splice:
$( [| "u" |] ) "u" $( lift "u" ) "u"
But if I replace a working version:
pr n = [| putStrLn ( $(lift( nameBase n ++ " = " )) ++ show $(varE n) ) |] ----- case (i) -----
by
pr n = [| putStrLn ( $([| (nameBase n) ++ " = " |]) ++ show $(varE n) ) |] ----- case (ii) -----
I again get the error
In the working version, 'n' appears inside a splice, whereas in the other n is in a quote. AFAIK any value can be used in a splice (provided it meets the staging restrictions), whereas only Lift-able values can be used in a quote.
If I take this as a granted axiom, then I can admit the behavior above (error in case (ii), whereas it is working in case (i)) because n is a (Name), and so is not instance of Lift. Thus we are compelled to use lift instead of [||] (although the behavior is about the same for both in simple examples, as shown in my example above for "u"). I do not understand the exact reason for that, but I can do without; and maybe it is better, because I am very probably not enough experienced to understand the details (and the reason is perhaps not trivial when I read Oleg who writes that what gives an error above in Haskell works in MetaOCaml). What is strange is that: * in the version using "lift", the definition of lift asks for the output of (nameBase n) to be an instance of Lift, what is the case because it is a string (cf my previous post in this thread). * whereas in the second version, we ask for n, not (nameBase n), to be an instance of Lift. Anyway, if we admit your axiom as granted, then we can also admit that the following version does not work (version of my initial post):
pr :: Name -> ExpQ pr n = [| putStrLn $ (nameBase n) ++ " = " ++ show $(varE n) |]
Thanks, TP
lift and [| |] give similar results for that very stripped-down example, but it would be incorrect to extrapolate their behaviors from that case. They're executed at different times, by different mechanisms, and have vastly different behavior. It's also best to think of [| |] as having the type String -> ExpQ, even though the input isn't syntactically quoted. Suppose you have a slightly different example:
let x = 1 :: Int runQ $ lift $ show x ListE [LitE (CharL '1')] runQ [| show x |] AppE (VarE GHC.Show.show) (VarE x_1627398832)
With lift, the expression is evaluated, then the result '1' is lifted into an AST. But TH quotes do something entirely different: they lift *the expression* into an AST. In order to do so, the quoting mechanism needs to parse its input string, then determine what each identifier is referring to. When you're defining a function:
let p :: (Show a, Lift a) => a -> ExpQ; p n = [| show n |]
The quote has two terms: show and n. 'n' is a lambda-bound value, and show is free. Free variables are looked up in the environment. That's why we see 'VarE GHC.Show.show' in the AST above; the fully-qualified Name is generated and the AST references that name. (numeric and string literals are represented directly) This is the key difference between this function definition and running the splice above: in the function 'n' is lambda-bound, whereas in the above splice 'x' is a free variable. Lambda bindings can't be referenced by name because that name may not be in scope when the generated splice is run. Instead, lambda-bound values must be lifted directly into the AST, which is exactly what 'lift' does. If we apply the function to a value, we can see the generated AST:
runQ (p 2) AppE (VarE GHC.Show.show) (LitE (IntegerL 2))
The generated AST has the number 2, and applies the function GHC.Show.show to it. If we want to show something that doesn't have a Lift instance, we can't do it directly. However, we can do this:
let q :: Show a => a -> ExpQ; q n = [| $(lift $ show n) |] runQ (q 2) ListE [LitE (CharL '2')]
Note the differences. We no longer require that 'n' has a Lift instance.
However, the actual value of 'n' never appears in the AST! Instead, we
first show 'n', then lift in the resulting string. The order of operations
is changed too. In the first case, the literal 2 is lifted into the AST
via lift, and the generated splice will apply show to that number whenever
the splice is run. In the second case, (show 2) is evaluated first, then
the result is lifted into the AST (again via lift), causing that string to
be referenced within the splice.
HTH,
John
On Wed, Jul 3, 2013 at 5:44 AM, TP
John Lato wrote:
Now, I have found another behavior difficult to understand for me:
runQ $ lift "u" ListE [LitE (CharL 'u') runQ $ [| "u" |] LitE (StringL "u")
So we have similar behaviors for lift and [||]. We can check it in a splice:
$( [| "u" |] ) "u" $( lift "u" ) "u"
But if I replace a working version:
pr n = [| putStrLn ( $(lift( nameBase n ++ " = " )) ++ show $(varE n) ) |] ----- case (i) -----
by
pr n = [| putStrLn ( $([| (nameBase n) ++ " = " |]) ++ show $(varE n) ) |] ----- case (ii) -----
I again get the error
In the working version, 'n' appears inside a splice, whereas in the other n is in a quote. AFAIK any value can be used in a splice (provided it meets the staging restrictions), whereas only Lift-able values can be used in a quote.
If I take this as a granted axiom, then I can admit the behavior above (error in case (ii), whereas it is working in case (i)) because n is a (Name), and so is not instance of Lift. Thus we are compelled to use lift instead of [||] (although the behavior is about the same for both in simple examples, as shown in my example above for "u").
I do not understand the exact reason for that, but I can do without; and maybe it is better, because I am very probably not enough experienced to understand the details (and the reason is perhaps not trivial when I read Oleg who writes that what gives an error above in Haskell works in MetaOCaml).
What is strange is that: * in the version using "lift", the definition of lift asks for the output of (nameBase n) to be an instance of Lift, what is the case because it is a string (cf my previous post in this thread). * whereas in the second version, we ask for n, not (nameBase n), to be an instance of Lift.
Anyway, if we admit your axiom as granted, then we can also admit that the following version does not work (version of my initial post):
pr :: Name -> ExpQ pr n = [| putStrLn $ (nameBase n) ++ " = " ++ show $(varE n) |]
Thanks,
TP
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participants (4)
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adam vogt -
John Lato -
oleg＠okmij.org -
TP