Calling @code_native on a function in a module segfaults.

[KristofferC]

This segfaults julia 0.5:

module Foo
    foo() = 1
end

# Note lack of parenthesis
@code_native Foo.foo

In 0.4.3 this gives:

ERROR: no method found for the specified argument types

[vtjnash]

it looks like the key failure here is that it tries to emit code for the builtin function Core.getfield(?), which has the following AST, so it fails in jl_new_lambda_info:

(lldb) p jl_(method->ast) Expr(:lambda)::Any

[rfourquet]

Is it the same problem as the segfault in @code_native applicable(typemax, Int) ?

[yuyichao]

Yes, applicable is also a builtin.

[mauro3]

Another one:

    _       _ _(_)_     |  A fresh approach to technical computing
  (_)     | (_) (_)    |  Documentation: http://docs.julialang.org
   _ _   _| |_  __ _   |  Type "?help" for help.
  | | | | | | |/ _` |  |
  | | |_| | | | (_| |  |  Version 0.5.0-pre+5609 (2016-07-22 07:00 UTC)
 _/ |\__'_|_|_|\__'_|  |  Commit 21f28c9* (0 days old master)
|__/                   |  x86_64-pc-linux-gnu

julia> first(take(1:Inf,2))

signal (11): Segmentation fault
while loading no file, in expression starting on line 0
jl_f_tuple at /home/mauro/julia/julialang_upstream/src/builtins.c:648
eval_user_input at ./REPL.jl:74
jl_call_method_internal at /home/mauro/julia/julialang_upstream/src/julia_internal.h:177 [inlined]
jl_apply_generic at /home/mauro/julia/julialang_upstream/src/gf.c:2034
macro expansion at ./REPL.jl:92 [inlined]
#3 at ./event.jl:46
jl_call_method_internal at /home/mauro/julia/julialang_upstream/src/julia_internal.h:177 [inlined]
jl_apply_generic at /home/mauro/julia/julialang_upstream/src/gf.c:2034
jl_apply at /home/mauro/julia/julialang_upstream/src/julia.h:1394 [inlined]
start_task at /home/mauro/julia/julialang_upstream/src/task.c:253
unknown function (ip: 0xffffffffffffffff)
Allocations: 1621003 (Pool: 1620110; Big: 893); GC: 0
zsh: segmentation fault (core dumped)  julia5

[mauro3]

I'm on a roll! This segfaults too f(aa) = (a,b=aa.a,aa.b; a+b) as does this mal-formed expression function f(aa)a,b=aa.a,aa.b; a+b).

[yuyichao]

Those are unrelated.

[mauro3]

You're sure? The backtrack is exactly the same though (modulo different allocation numbers). Should I open another issue?

[yuyichao]

You're sure?

Yes.

Should I open another issue?

No, #17548

[mauro3]

Thanks. No wonder it was so easy to find other examples...

[yuyichao]

The backtrack is exactly the same though (modulo different allocation numbers)

Also note that both the backtrace and the allocation number are useless in terms of directly identify the cause of the failure. They are only useful when debugging the issue.