Accessing to the AST of a function in Julia - julia

In Julia, it is possible to view the AST of a user defined function:
julia> myFunc(x) = 5*x+3
myFunc (generic function with 1 method)
julia> tmp = dump(quote myFunc end)
Expr
head: Symbol block
args: Array(Any,(2,))
1: Expr
head: Symbol line
args: Array(Any,(2,))
1: Int64 1
2: Symbol none
typ: Any
2: Symbol myFunc
typ: Any
Which is the AST I am interested in. However, the variable tmp doesn't contain the Expr representing the syntactic tree I am expecting:
julia> tmp
julia> typeof(tmp)
Nothing (constructor with 1 method)
Is there another way to get this Expr? (the one that is displayed when running dump(quote myFunc end) )

dump does not give you the result; it is just a way of printing the value. (As you saw, it prints as a side-effect and returns a nothing.)
What you gave dump was an AST containing the name of your function, not the function itself. dump is not printing out a representation of your function: it is saying it has a block of one line containing the symbol myFunc.
If you want the AST, you should run code_typed(myFunc,(Any,)) or code_lowered(myFunc,(Any,)). For other functions, you will need different and/or more specific type signatures as the second argument.
If you are only planning to call myFunc with Ints or Float64s or whatever, use that instead of Any -- it will make a difference to code_typed's output, since the type inference will change.
I wrote a blog post documenting the code_typed/code_lowered set of functions: http://blog.leahhanson.us/julia-introspects.html
(I also spend time in that post looking at their output, the Expr type and explaining it's structure.)
The Metaprogramming section of the official manual will probably be useful to you in working with ASTs, if you haven't already read it.

You can't access the AST of a function, because a function is a a collection of methods (that might be implemented differently) in Julia. If it suits your needs you should use the documented code_typed function, where you specify the types of the arguments to select the right method. There are also some hints in Access the AST for generic functions in Julia, but that is not documented functionality, so it might change without warning.

Related

Julia: Even-number datatype for functions

I have about 50 functions which should consume only even positive numbers. Right now I am checking each time with an "if" whether the number put in is zero or not:
function grof(x::Int)
if (x % 2) == 0
println("good")
else
throw("x is not an even number!!!!!!!!!!!!! Stupid programmer!")
end
end
Ideally, I would like to have a datatype which produces this automatically, i.e.
function grof(x::EvenInt)
println("good")
end
However, I am not able to produce this datatype by my own since I am unable to understand the documentary. Thanks for your help!
Best, v.
I don't think creating a type is warranted in such a situation: I would simply #assert that the condition is verified at the beginning of the function(s). (Funnily enough, checking the whether a number is even is the example that was chosen in the documentation to illustrate the effect of #assert)
For example:
julia> function grof(x::Int)
#assert iseven(x) "Stupid programmer!"
println("good")
end
grof (generic function with 1 method)
julia> grof(2)
good
julia> grof(3)
ERROR: AssertionError: Stupid programmer!
Stacktrace:
[1] grof(::Int64) at ./REPL[5]:2
[2] top-level scope at REPL[7]:1
EDIT: If you really want to create a type enforcing such a constraint, it is possible. The way to do this would be to
create a type (possibly subtyping one of the Number abstract types; maybe Signed)
define an inner constructor ensuring that such a type cannot hold an odd value
A very simple example to build upon would be along the lines of:
# A wrapper around an even integer value
struct EvenInt
val :: Int
# inner constructor
function EvenInt(val)
#assert iseven(val)
new(val)
end
end
# Accessor to the value of an EvenInt
val(x::EvenInt) = x.val
# A method working only on even numbers
grof(x::EvenInt) = println("good: $(val(x)) is even")
You'd use this like so:
julia> x = EvenInt(42)
EvenInt(42)
julia> grof(x)
good: 42 is even
julia> y = EvenInt(1)
ERROR: AssertionError: iseven(val)
Stacktrace:
[1] EvenInt(::Int64) at ./REPL[1]:5
[2] top-level scope at REPL[6]:1
but note that you can't do anything on EvenInts yet: you need to either unwrap them (using val() in this case), or define operations on them (a task which can be vastly simplified if you make EvenInt a subtype of one of the abstract number types and follow the relevant interface).
All integers multiplied by two are even, so redefine your function to take half the number it currently takes.
function grof2(halfx::Int)
x=2*halfx
println("good")
end

Why is a Julia function name (without arguments) silently ignored?

I have a script that defines a function, and later intended to call the function but forgot to add the parentheses, like this:
function myfunc()
println("inside myfunc")
end
myfunc # This line is silently ignored. The function isn't invoked and there's no error message.
After a while I did figure out that I was missing the parentheses, but since Julia didn't give me an error, I'm wondering what that line is actually doing? I'm assuming that it must be doing something with the myfunc statement, but I don't know Julia well enough to understand what is happening.
I tried --depwarn=yes but don't see a julia command line switch to increase the warning level or verbosity. Please let me know if one exists.
For background context, the reason this came up is that I'm trying to translate a Bash script to Julia, and there are numerous locations where an argument-less function is defined and then invoked, and in Bash you don't need parentheses after the function name to invoke it.
The script is being run from command line (julia stub.jl) and I'm using Julia 1.0.3 on macOS.
It doesn't silently ignore the function. Calling myfunc in an interactive session will show you what happens: the call returns the function object to the console, and thus call's the show method for Function, showing how many methods are currently defined for that function in your workspace.
julia> function myfunc()
println("inside myfunc")
end
myfunc (generic function with 1 method)
julia> myfunc
myfunc (generic function with 1 method)
Since you're calling this in a script, show is never called, and thus you don't see any result. But it doesn't error, because the syntax is valid.
Thanks to DNF for the helpful comment on it being in a script.
It does nothing.
As in c, an expression has a value: in c the expression _ a=1+1; _ has the value _ 2 _ In c, this just fell out of the parser: they wanted to be able to evaluate expressions like _ a==b _
In Julia, it's the result of designing a language where the code you write is handled as a data object of the language. In Julia, the expression "a=1+1" has the value "a=1+1".
In c, the fact that
a=1+1;
is an acceptable line of code means that, accidentally,
a;
is also an acceptable line of code. The same is true in Julia: the compiler expects to see a data value there: any data value you put may be acceptable: even for example the data value that represents the calculated value returned by a function:
myfunc()
or the value that represents the function object itself:
myfunc
As in c, the fact that data values are everywhere in your code just indicates that the syntax allows data values everywhere in your code and that the compiler does nothing with data values that are everywhere in your code.

What does the "Base" keyword mean in Julia?

I saw this example in the Julia language documentation. It uses something called Base. What is this Base?
immutable Squares
count::Int
end
Base.start(::Squares) = 1
Base.next(S::Squares, state) = (state*state, state+1)
Base.done(S::Squares, s) = s > S.count;
Base.eltype(::Type{Squares}) = Int # Note that this is defined for the type
Base.length(S::Squares) = S.count;
Base is a module which defines many of the functions, types and macros used in the Julia language. You can view the files for everything it contains here or call whos(Base) to print a list.
In fact, these functions and types (which include things like sum and Int) are so fundamental to the language that they are included in Julia's top-level scope by default.
This means that we can just use sum instead of Base.sum every time we want to use that particular function. Both names refer to the same thing:
Julia> sum === Base.sum
true
Julia> #which sum # show where the name is defined
Base
So why, you might ask, is it necessary is write things like Base.start instead of simply start?
The point is that start is just a name. We are free to rebind names in the top-level scope to anything we like. For instance start = 0 will rebind the name 'start' to the integer 0 (so that it no longer refers to Base.start).
Concentrating now on the specific example in docs, if we simply wrote start(::Squares) = 1, then we find that we have created a new function with 1 method:
Julia> start
start (generic function with 1 method)
But Julia's iterator interface (invoked using the for loop) requires us to add the new method to Base.start! We haven't done this and so we get an error if we try to iterate:
julia> for i in Squares(7)
println(i)
end
ERROR: MethodError: no method matching start(::Squares)
By updating the Base.start function instead by writing Base.start(::Squares) = 1, the iterator interface can use the method for the Squares type and iteration will work as we expect (as long as Base.done and Base.next are also extended for this type).
I'll grant that for something so fundamental, the explanation is buried a bit far down in the documentation, but http://docs.julialang.org/en/release-0.4/manual/modules/#standard-modules describes this:
There are three important standard modules: Main, Core, and Base.
Base is the standard library (the contents of base/). All modules
implicitly contain using Base, since this is needed in the vast
majority of cases.

Julia macro doesn't recognize Array argument

I wrote a fairly complex macro that manipulates an array, but it didn't work. I get the same error message even after replacing my macro with something ridiculously simple:
macro len(arr::Array)
length(arr)
end
#len [1 2 3]
The error message I get is:
ERROR: LoadError: MethodError: no method matching #len(::Expr)
Why is Julia insisting on interpreting my input as an expression? I'm completely new to Julia, and there is obviously something I don't understand about either macros or the type system. Can someone please explain?
EDIT: Time to explain what my original macro was for. I want to take a table as input and use it to define variables in the local scope. The entries in the first column define individual variable names and the other columns define variable contents. Since a function can't do this, I need a macro. I have a working macro that takes a long (triple-quoted) string as input, but now I want it to take an array as input instead.
After this explanation, maybe I should add another subquestion: why does my other macro accept the appropriate input (it is defined as macro foo(text::String)) whereas the array version doesn't?
User Chris Rackauckas answered this on irc in discussion with me:
#len $a
macros are on expressions, not on the values, so
this sounds like something you'd want a function for
(if it's really needed at all).
I'm not sure this would work there anyway.
This isn't what macros are for..
it's for converting expressions into other expressions, mostly.
If you're trying to use macros on values, then you want a function
(you may be able to #eval or something to make this work here, but you'll be going through great lengths to make a macro act like a function)
..as a function:
julia> function len(x)
length(x)
end
len (generic function with 1 method)
julia> len(a)
3
julia> a
1x3 Array{Int64,2}:
1 2 3
To answer my own question: this is how that "ridiculously simple" macro should have been written.
macro len(expr)
:(length($expr))
end
This operates on an expression, not an array (thank you Chris Rackauckas and Isaiah for pointing that out). I was misled by my own experiments, since a similar macro for strings happens to work:
# MISLEADING CODE, DON'T USE THIS!
macro lenstr(str::String)
length(str)
end
Apparently Julia allows string macros, but this is an exception to the general rule that macros operate on expressions. To illustrate how arguments appear to the macro:
macro dump(arg)
dump(arg)
end
julia> #dump "foo bar"
String "foo bar"
julia> #dump [1,2,3]
Expr
head: Symbol vect
args: Array{Any}((3,))
1: Int64 1
2: Int64 2
3: Int64 3
typ: Any
julia> #dump [1 2 3]
Expr
head: Symbol hcat
args: Array{Any}((3,))
1: Int64 1
2: Int64 2
3: Int64 3
typ: Any
Note how [1,2,3] and [1 2 3] are seen by the macro as expressions, not arrays. This was the source of my error. By applying lessons learned here, my more complex macro that creates variables in a local scope from an array now works. Thank you all!

julia introspection - get name of variable passed to function

In Julia, is there any way to get the name of a passed to a function?
x = 10
function myfunc(a)
# do something here
end
assert(myfunc(x) == "x")
Do I need to use macros or is there a native method that provides introspection?
You can grab the variable name with a macro:
julia> macro mymacro(arg)
string(arg)
end
julia> #mymacro(x)
"x"
julia> #assert(#mymacro(x) == "x")
but as others have said, I'm not sure why you'd need that.
Macros operate on the AST (code tree) during compile time, and the x is passed into the macro as the Symbol :x. You can turn a Symbol into a string and vice versa. Macros replace code with code, so the #mymacro(x) is simply pulled out and replaced with string(:x).
Ok, contradicting myself: technically this is possible in a very hacky way, under one (fairly limiting) condition: the function name must have only one method signature. The idea is very similar the answers to such questions for Python. Before the demo, I must emphasize that these are internal compiler details and are subject to change. Briefly:
julia> function foo(x)
bt = backtrace()
fobj = eval(current_module(), symbol(Profile.lookup(bt[3]).func))
Base.arg_decl_parts(fobj.env.defs)[2][1][1]
end
foo (generic function with 1 method)
julia> foo(1)
"x"
Let me re-emphasize that this is a bad idea, and should not be used for anything! (well, except for backtrace display). This is basically "stupid compiler tricks", but I'm showing it because it can be kind of educational to play with these objects, and the explanation does lead to a more useful answer to the clarifying comment by #ejang.
Explanation:
bt = backtrace() generates a ... backtrace ... from the current position. bt is an array of pointers, where each pointer is the address of a frame in the current call stack.
Profile.lookup(bt[3]) returns a LineInfo object with the function name (and several other details about each frame). Note that bt[1] and bt[2] are in the backtrace-generation function itself, so we need to go further up the stack to get the caller.
Profile.lookup(...).func returns the function name (the symbol :foo)
eval(current_module(), Profile.lookup(...)) returns the function object associated with the name :foo in the current_module(). If we modify the definition of function foo to return fobj, then note the equivalence to the foo object in the REPL:
julia> function foo(x)
bt = backtrace()
fobj = eval(current_module(), symbol(Profile.lookup(bt[3]).func))
end
foo (generic function with 1 method)
julia> foo(1) == foo
true
fobj.env.defs returns the first Method entry from the MethodTable for foo/fobj
Base.decl_arg_parts is a helper function (defined in methodshow.jl) that extracts argument information from a given Method.
the rest of the indexing drills down to the name of the argument.
Regarding the restriction that the function have only one method signature, the reason is that multiple signatures will all be listed (see defs.next) in the MethodTable. As far as I know there is no currently exposed interface to get the specific method associated with a given frame address. (as an exercise for the advanced reader: one way to do this would be to modify the address lookup functionality in jl_getFunctionInfo to also return the mangled function name, which could then be re-associated with the specific method invocation; however, I don't think we currently store a reverse mapping from mangled name -> Method).
Note also that (1) backtraces are slow (2) there is no notion of "function-local" eval in Julia, so even if one has the variable name, I believe it would be impossible to actually access the variable (and the compiler may completely elide local variables, unused or otherwise, put them in a register, etc.)
As for the IDE-style introspection use mentioned in the comments: foo.env.defs as shown above is one place to start for "object introspection". From the debugging side, Gallium.jl can inspect DWARF local variable info in a given frame. Finally, JuliaParser.jl is a pure-Julia implementation of the Julia parser that is actively used in several IDEs to introspect code blocks at a high level.
Another method is to use the function's vinfo. Here is an example:
function test(argx::Int64)
vinfo = code_lowered(test,(Int64,))
string(vinfo[1].args[1][1])
end
test (generic function with 1 method)
julia> test(10)
"argx"
The above depends on knowing the signature of the function, but this is a non-issue if it is coded within the function itself (otherwise some macro magic could be needed).

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