preop expression and variable

preop expression and variable

[Guillaume Oberlé]

Hi,

I'm an intern and I'm working on a project that aims to use sparse to generate a Json AST of an OpenCL code. Inspired by ast-inspect.c, I've started writing a C to Json program for sparse, and I have some issues.

Here is the code : https://gist.github.com/goberle/b867d29c1b833cf55d36 and here is an example of the output : https://gist.github.com/goberle/e3e8e35bf6a05ea86a6d.

If you look at the output (l117 or l133), I have a PreopExpression before every variable in an expression statement, even if I have no preoperator. I wonder if these are usefull, and if so, what they represent ? (EXPR_PREOP case is defined at line 443 in the code)

Thank you for your time and feel free to contact me if you have any other suggestions.

goberle

Re: preop expression and variable

[Linus Torvalds]

On Fri, Mar 21, 2014 at 4:02 AM, Guillaume Oberlé <goberle@ygg.tf> wrote:
>
> If you look at the output (l117 or l133), I have a PreopExpression before every variable in an expression statement, even if I have no preoperator.

You need to look at what the op is too, not just the type.

I assume you're talking about EXPR_PREOP-'*'. That's a dereference
("lvalue access").  And yes, it's required. A symbol on its own is
just the address or the symbol. Actually *accessing* that symbol is
that PREOP-*.

            Lnus
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