I have been messing around with generated functions in Julia, and have come to a weird problem I do not understand fully: My final goal would involve calling a macro (more specifically #tullio) from within a generated function (to perform some tensor contractions that depend on the input tensors). But I have been having problems, which I narrowed down to calling the macro from within the generated function.
To illustrate the problem, let's consider a very simple example that also fails:
macro my_add(a,b)
return :($a + $b)
end
function add_one_expr(x::T) where T
y = one(T)
return :( #my_add($x,$y) )
end
#generated function add_one_gen(x::T) where T
y = one(T)
return :( #my_add($x,$y) )
end
With these declarations, I find that eval(add_one_expr(2.0)) works just as expected and returns and expression
:(#my_add 2.0 1.0)
which correctly evaluates to 3.0.
However evaluating add_one_gen(2.0) returns the following error:
MethodError: no method matching +(::Type{Float64}, ::Float64)
Doing some research, I have found that #generated actually produces two codes, and in one only the types of the variables can be used. I think this is what is happening here, but I do not understand what is happening at all. It must be some weird interaction between macros and generated functions.
Can someone explain and/or propose a solution? Thank you!
I find it helpful to think of generated functions as having two components: the body and any generated code (the stuff inside a quote..end). The body is evaluated at compile time, and doesn't "know" the values, only the types. So for a generated function taking x::T as an argument, any references to x in the body will actually point to the type T. This can be very confusing. To make things clearer, I recommend the body only refer to types, never to values.
Here's a little example:
julia> #generated function show_val_and_type(x::T) where {T}
quote
println("x is ", x)
println("\$x is ", $x)
println("T is ", T)
println("\$T is ", $T)
end
end
show_val_and_type
julia> show_val_and_type(3)
x is 3
$x is Int64
T is Int64
$T is Int64
The interpolated $x means "take the x from the body (which refers to T) and splice it in.
If you follow the approach of never referring to values in the body, you can test generated functions by removing the #generated, like this:
julia> function add_one_gen(x::T) where T
y = one(T)
quote
#my_add(x,$y)
end
end
add_one_gen
julia> add_one_gen(3)
quote
#= REPL[42]:4 =#
#= REPL[42]:4 =# #my_add x 1
end
That looks reasonable, but when we test it we get
julia> add_one_gen(3)
ERROR: UndefVarError: x not defined
Stacktrace:
[1] macro expansion
# ./REPL[48]:4 [inlined]
[2] add_one_gen(x::Int64)
# Main ./REPL[48]:1
[3] top-level scope
# REPL[49]:1
So let's see what the macro gives us
julia> #macroexpand #my_add x 1
:(Main.x + 1)
It's pointing to Main.x, which doesn't exist. The macro is being too eager, and we need to delay its evaluation. The standard way to do this is with esc. So finally, this works:
julia> macro my_add(a,b)
return :($(esc(a)) + $(esc(b)))
end
#my_add
julia> #generated function add_one_gen(x::T) where T
y = one(T)
quote
#my_add(x,$y)
end
end
add_one_gen
julia> add_one_gen(3)
4
Related
I am fairly new to Julia and I am learning about metaprogramming.
I would like to write a macro that receive in input a function and returns another function based on the implementation details of its input.
For example given:
function f(x)
x + 100
end
function g(x)
f(x)*x
end
function h(x)
g(x)-0.5*f(x)
end
I would like to write a macro that returns something like that:
function h_traced(x)
f = x + 100
println("loc 1 x: ", x)
g = f * x
println("loc 2 x: ", x)
res = g - 0.5 * f
println("loc 3 x: ", x)
Now both code_lowered and code_typed seems to give me back the AST in the form of CodeInfo, however when I try to use it programmatically in my macro I get empty object.
macro myExpand(f)
body = code_lowered(f)
println("myExpand Body lenght: ",length(body))
end
called like this
#myExpand :(h)
however the same call outside the macro works ok.
code_lowered(h)
At last even the following return an empty CodeInfo.
macro myExpand(f)
body = code_lowered(Symbol("h"))
println("myExpand Body lenght: ",length(body))
end
This might be incredible trivial but I could not work out myseld why the h symbol does not resolve to the function defined. Am I missing something about the scope of symbols?
I find it useful to think about macros as a way to transform an input syntax into an output syntax.
So you could very well define a macro #my_macro such that
#my_macro function h(x)
g(x)-0.5*f(x)
end
would expand to something like
function h_traced(x)
println("entering function: x=", x)
g(x)-0.5*f(x)
end
But to such a macro, h is merely a name, an identifier (technically, a Symbol) that can be transformed into h_traced. h is not the function that is bound to this name (in the same way as x = 2 involves binding a name x, to an integer value 2, but x is not 2; x is merely a name that can be used to refer to 2). In contrast to this, when you call code_lowered(h), h gets evaluated first, and code_lowered is passed its value (which is a function) as argument.
Back to our macro: expanding to an expression that involves the definition of g and f goes way further than mere syntax transformations: we're leaving the purely syntactic domain, since such a transformation would need to "understand" that these are functions, look up their definitions and so on.
You are right to think about code_lowered and friends: this is IMO the adequate level of abstraction for what you're trying to achieve. You should probably look into tools like Cassette.jl or IRTools.jl. That being said, if you're still relatively new to Julia, you might want to get a bit more used to the language before delving too deeply into such topics.
You don't need a macro, you need a generated function. They can not only return code (Expr), but also IR (lowered code). Usually, for this kind of thing, people use Base.uncompressed_ast, not code_lowered. Both Cassette and IRTools simplify the implementation for you, in different ways.
The basic idea is:
Have a generated function that takes a function and its arguments
In that function, get the IR of that function, and modify it to your purposes
Return the new IR from the generated function. This will then be compiled and called on the original arguments.
A short demonstration with IRTools:
julia> IRTools.#dynamo function traced(args...)
ir = IRTools.IR(args...)
p = IRTools.Pipe(ir)
for (v, stmt) in p
IRTools.insertafter!(p, v, IRTools.xcall(println, "loc $v"))
end
return IRTools.finish(p)
end
julia> function h(x)
sin(x)-0.5*cos(x)
end
h (generic function with 1 method)
julia> #code_ir traced(h, 1)
1: (%1, %2)
%3 = Base.getfield(%2, 1)
%4 = Base.getfield(%2, 2)
%5 = Main.sin(%4)
%6 = (println)("loc %3")
%7 = Main.cos(%4)
%8 = (println)("loc %4")
%9 = 0.5 * %7
%10 = (println)("loc %5")
%11 = %5 - %9
%12 = (println)("loc %6")
return %11
julia> traced(h, 1)
loc %3
loc %4
loc %5
loc %6
0.5713198318738266
The rest is left as an exercise. The numbers of the variables are off, because they are, of course, shifted during the transformation. You'd have to add some bookkeeping for that, or use the substitute function on Pipe in some way (but I never quite understood it). If you need the name of the variables, you can get the IR with slots preserved by using a different method of the IR constructor.
(And now the advertisement: I have written something like this. It's currently quite inefficient, but you might get some ideas from it.)
I am a noob at metaprogramming so maybe I am not understanding this. I thought the purpose of the #nloops macro in Base.Cartesian was to make it possible to code an arbitrary number of nested for loops, in circumstances where the dimension is unknown a priori. In the documentation for the module, the following example is given:
#nloops 3 i A begin
s += #nref 3 A i
end
which evaluates to
for i_3 = 1:size(A,3)
for i_2 = 1:size(A,2)
for i_1 = 1:size(A,1)
s += A[i_1,i_2,i_3]
end
end
end
Here, the number 3 is known a priori. For my purposes, however, and for the purposes that I thought nloops was created, the number of nested levels is not known ahead of time. So I would not be able to hard code the integer 3. Even in the documentation, it is stated:
The (basic) syntax of #nloops is as follows:
The first argument must be an integer (not a variable) specifying the number of loops.
...
If I assign an integer value - say the dimension of an array that is passed to a function - to some variable, the nloops macro no longer works:
b = 3
#nloops b i A begin
s += #nref b A i
end
This returns an error:
ERROR: LoadError: MethodError: no method matching _nloops(::Symbol, ::Symbol, ::Symbol, ::Expr)
Closest candidates are:
_nloops(::Int64, ::Symbol, ::Symbol, ::Expr...) at cartesian.jl:43
...
I don't know how to have nloops evaluate the b variable as an integer rather than a symbol. I have looked at the documentation and tried various iterations of eval and other functions and macros but it is either interpreted as a symbol or an Expr. What is the correct, julian way to write this?
See supplying the number of expressions:
julia> A = rand(4, 4, 3) # 3D array (Array{Int, 3})
A generated function is kinda like a macro, in that the resulting expression is not returned, but compiled and executed on invocation/call, it also sees the type (and their type parameters of course) of the arguments, ie:
inside the generated function, A is Array{T, N}, not the value of the array.
so T is Int and N is 3!
Here inside the quoted expression, N is interpolated into the expression, with the syntax $N, which evaluates to 3:
julia> #generated function mysum(A::Array{T,N}) where {T,N}
quote
s = zero(T)
#nloops $N i A begin
s += #nref $N A i
end
s
end
end
mysum (generic function with 1 method)
julia> mysum(A)
23.2791638775186
You could construct the expression and then evaluate it, ie.:
julia> s = 0; n = 3;
julia> _3loops = quote
#nloops $n i A begin
global s += #nref $n A i
end
end
quote
#nloops 3 i A begin
global s += #nref(3, A, i)
end
end
julia> eval(_3loops)
julia> s
23.2791638775186
I have scrubbed manually the LineNumberNodes from the AST for readability (there is also MacroTools.prettify, that does it for you).
Running this example in the REPL needs to declare s as global inside the loop in Julia 1.0.
I'd like to write a simple macro that shows the names & values of variables. In Common Lisp it would be
(defmacro dprint (&rest vars)
`(progn
,#(loop for v in vars
collect `(format t "~a: ~a~%" ',v ,v))))
In Julia I had two problems writing this:
How can I collect the generated Expr objects into a block? (In Lisp, this is done by splicing the list with ,# into progn.) The best I could come up with is to create an Expr(:block), and set its args to the list, but this is far from elegant.
I need to use both the name and the value of the variable. Interpolation inside strings and quoted expressions both use $, which complicates the issue, but even if I use string for concatenation, I can 't print the variable's name - at least :($v) does not do the same as ',v in CL...
My current macro looks like this:
macro dprint(vars...)
ex = Expr(:block)
ex.args = [:(println(string(:($v), " = ", $v))) for v in vars]
ex
end
Looking at a macroexpansion shows the problem:
julia> macroexpand(:(#dprint x y))
quote
println(string(v," = ",x))
println(string(v," = ",y))
end
I would like to get
quote
println(string(:x," = ",x))
println(string(:y," = ",y))
end
Any hints?
EDIT: Combining the answers, the solution seems to be the following:
macro dprint(vars...)
quote
$([:(println(string($(Meta.quot(v)), " = ", $v))) for v in vars]...)
end
end
... i.e., using $(Meta.quot(v)) to the effect of ',v, and $(expr...) for ,#expr. Thank you again!
the #show macro already exists for this. It is helpful to be able to implement it yourself, so later you can do other likes like make one that will show the size of an Array..
For your particular variant:
Answer is Meta.quot,
macro dprint(vars...)
ex = Expr(:block)
ex.args = [:(println($(Meta.quot(v)), " = ", $v)) for v in vars]
ex
end
See with:
julia> a=2; b=3;
julia> #dprint a
a = 2
julia> #dprint a b
a = 2
b = 3
oxinabox's answer is good, but I should mention the equivalent to ,#x is $(x...) (this is the other part of your question).
For instance, consider the macro
macro _begin(); esc(:begin); end
macro #_begin()(args...)
quote
$(args...)
end |> esc
end
and invocation
#begin x=1 y=2 x*y
which (though dubiously readable) produces the expected result 2. (The #_begin macro is not part of the example; it is required however because begin is a reserved word, so one needs a macro to access the symbol directly.)
Note
julia> macroexpand(:(#begin 1 2 3))
quote # REPL[1], line 5:
1
2
3
end
I consider this more readable, personally, than pushing to the .args array.
This question builds off of a previous SO question which was for building expressions from expressions inside of of a macro. However, things got a little trucker when quoting the whole expression. For example, I want to build the expression :(name=val). The following:
macro quotetest(name,val)
quote
nm = Meta.quot($(QuoteNode(name)))
v = Meta.quot($(QuoteNode(val)))
println(nm); println(typeof(nm))
println(v); println(typeof(val))
end
end
#quotetest x 5 # Test case: build :(x=5)
prints out
:x
Expr
$(Expr(:quote, 5))
Expr
showing that I am on the right path: nm and val are the expressions that I want inside of the quote. However, I can't seem to apply the previous solution at this point. For example,
macro quotetest(name,val)
quote
nm = Meta.quot($(QuoteNode(name)))
v = Meta.quot($(QuoteNode(val)))
println(nm); println(typeof(nm))
println(v); println(typeof(v))
println(:($(Expr(:(=),$(QuoteNode(nm)),$(QuoteNode(val))))))
end
end
fails, saying nm is not defined. I tried just interpolating without the QuoteNode, escaping the interpolation $(esc(nm)), etc. I can't seem to find out how to make it build the expression.
I think you are using $ signs more than you need to. Is this what you're looking for?
julia> macro quotetest(name,val)
quote
expr = :($$(QuoteNode(name)) = $$(QuoteNode(val)))
println(expr)
display(expr)
println(typeof(expr))
end
end
#quotetest (macro with 1 method)
julia> #quotetest test 1
test = 1
:(test = 1)
Expr
I would like to inject code into a function. For concreteness, consider a simple simulater:
function simulation(A, x)
for t in 1:1000
z = randn(3)
x = A*x + z
end
end
Sometimes I would like to record the values of x every ten time-steps, sometimes the values of z every 20 time-steps, and sometimes I don't want to record any values. I could, of course, put some flags as arguments to the function, and have some if-else statements. But I would like to rather keep the simulation code clean, and only inject a piece of code like
if t%10 == 0
append!(rec_z, z)
end
into particular places of the function whenever I need it. For that, I'd like to write a macro such that monitoring a particular value becomes
#monitor(:z, 10)
simulation(A, x)
Is that possible with Julia's Metaprogramming capabilities?
No, you cannot use metaprogramming to inject code into an already-written function. Metaprogramming can only do things that you could directly write yourself at precisely the location where the macro itself is written. That means that a statement like:
#monitor(:z, 10); simulation(A, x)
cannot even modify the simulation(A, x) function call. It can only expand out to some normal Julia code that runs before simulation is called. You could, perhaps, include the simulation function call as an argument to the macro, e.g., #monitor(:z, 10, simulation(A, x)), but now all the macro can do is change the function call itself. It still cannot "go back" and add new code to a function that was already written.
You could, however, carefully and meticulously craft a macro that takes the function definition body and modifies it to add your debug code, e.g.,
#monitor(:z, 10, function simulation(A, x)
for t in 1:1000
# ...
end
end)
But now you must write code in the macro that traverses the code in the function body, and injects your debug statement at the correct place. This is not an easy task. And it's even harder to write in a robust manner that wouldn't break the moment you modified your actual simulation code.
Traversing code and inserting it is a much easier task for you to do yourself with an editor. A common idiom for debugging statements is to use a one-liner, like this:
const debug = false
function simulation (A, x)
for t in 1:1000
z = rand(3)
x = A*x + z
debug && t%10==0 && append!(rec_z, z)
end
end
What's really cool here is that by marking debug as constant, Julia is able to completely optimize away the debugging code when it's false — it doesn't even appear in the generated code! So there is no overhead when you're not debugging. It does mean, however, that you have to restart Julia (or reload the module it's in) for you to change the debug flag. Even when debug isn't marked as const, I cannot measure any overhead for this simple loop. And chances are, your loop will be more complicated than this one. So don't worry about performance here until you actually double-check that it's having an effect.
You might be interested in this which i just whipped up. It doesn't QUITE do what you are doing, but it's close. Generally safe and consistent places to add code are the beginning and end of code blocks. These macros allow you to inject some code in those location (and even pass code parameters!)
Should be useful for say toggle-able input checking.
#cleaninject.jl
#cleanly injects some code into the AST of a function.
function code_to_inject()
println("this code is injected")
end
function code_to_inject(a,b)
println("injected code handles $a and $b")
end
macro inject_code_prepend(f)
#make sure this macro precedes a function definition.
isa(f, Expr) || error("checkable macro must precede a function definition")
(f.head == :function) || error("checkable macro must precede a function definition")
#be lazy and let the parser do the hard work.
b2 = parse("code_to_inject()")
#inject the generated code into the AST.
unshift!(f.args[2].args, b2)
#return the escaped function to the parser so that it generates the new function.
return Expr(:escape, f)
end
macro inject_code_append(f)
#make sure this macro precedes a function definition.
isa(f, Expr) || error("checkable macro must precede a function definition")
(f.head == :function) || error("checkable macro must precede a function definition")
#be lazy and let the parser do the hard work.
b2 = parse("code_to_inject()")
#inject the generated code into the AST.
push!(f.args[2].args, b2)
#return the escaped function to the parser so that it generates the new function.
return Expr(:escape, f)
end
macro inject_code_with_args(f)
#make sure this macro precedes a function definition.
isa(f, Expr) || error("checkable macro must precede a function definition")
(f.head == :function) || error("checkable macro must precede a function definition")
#be lazy and let the parser do the hard work.
b2 = parse(string("code_to_inject(", join(f.args[1].args[2:end], ","), ")"))
#inject the generated code into the AST.
unshift!(f.args[2].args, b2)
#return the escaped function to the parser so that it generates the new function.
return Expr(:escape, f)
end
################################################################################
# RESULTS
#=
julia> #inject_code_prepend function p()
println("victim function")
end
p (generic function with 1 method)
julia> p()
this code is injected
victim function
julia> #inject_code_append function p()
println("victim function")
end
p (generic function with 1 method)
julia> p()
victim function
this code is injected
julia> #inject_code_with_args function p(a, b)
println("victim called with $a and $b")
end
p (generic function with 2 methods)
julia> p(1, 2)
injected code handles 1 and 2
victim called with 1 and 2
=#