From: John Max Skaller <skaller@ozemail.com.au>
To: Xavier Leroy <xavier.leroy@inria.fr>
Cc: caml-list@inria.fr
Subject: Re: [Caml-list] generating a call-graph
Date: Tue, 03 Jun 2003 14:27:43 +1000 [thread overview]
Message-ID: <3EDC23BF.2090902@ozemail.com.au> (raw)
In-Reply-To: <20030526111416.A31160@pauillac.inria.fr>
Xavier Leroy wrote:
>>Does anyone know a way of generating a call-graph from a set of ocaml
>>sources? What I want to do is, at a minimum, get a list of all the
>>functions that could be called as a result of a given function invocation.
>>
>
> This requires a non-trivial static analysis called "control flow
> analysis" in the literature; particular instances include Shivers'
> 0-CFA and k-CFA, Jagannathan and Wright's "polymorphic splitting",
> etc.
>
> The difficulty is that functions are first-class values
.. and the difficulty can be bypassed by considering
the actual industrial requirement, which probably
isn't as stated above.
We often want to know what the definition
of a function depends on, and that clearly *excludes*
any functions passed in as parameters.
Second, an incomplete graph would still be very useful.
For example to determine if you can move a function
definition earlier in the code, so as to call it from
some other function -- or whether you would have to move
a lot more functions back -- and if so which ones --
and, indeed, if it is possible at all (without recursion).
I guess a patch to the parser could gather the required
information easily ( the main difficulty being
the huge number of anonymous functions that tend to
float around when one does currying).
--
John Max Skaller, mailto:skaller@ozemail.com.au
snail:10/1 Toxteth Rd, Glebe, NSW 2037, Australia.
voice:61-2-9660-0850
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next prev parent reply other threads:[~2003-06-03 4:27 UTC|newest]
Thread overview: 5+ messages / expand[flat|nested] mbox.gz Atom feed top
2003-05-23 16:34 Yaron M. Minsky
2003-05-26 0:51 ` Jeff Henrikson
2003-05-26 9:14 ` Xavier Leroy
2003-06-03 4:27 ` John Max Skaller [this message]
2003-05-27 2:40 ` Jeff Henrikson
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